START 100/ch1and2a.tex part 1

Calulus pre-requisites you must know.
  
Derivative = slope of tangent line = rate.

Integral = area between curve and x-axis (where area can be negative).

---

The Fundamental Theorem of Calculus:  Suppose \(f\) continuous on \([a, b]\).
   
1.)  If \(G(x) = \int_a^x f(t) dt\), then \(G'(x) = f(x)\).
I.e.,  \({d \over dx[ \int_a^x f(t) dt] = f(x)\).}
 
2.)   \(\int_a^b f(t) dt = F(b) - F(a)\) where \(F\) is any antiderivative of 
\(f\), that is \(F' = f\).

---

Integration Pre-requisites:

* Integration by substitution

 * Integration by parts

* Integration by partial fractions

Note: Partial fractions are also used in ch 6 for a different application.

---

Suppose \(f\) is cont. on \((a, b)\) and the
point \(t_0 \in (a, b)\), 

Solve IVP:~~~~~~~~ \({dy \over dt} = f(t)\),~~~~~
 \(y(t_0) = y_0\)

\(dy = f(t)dt\)~~~~~~
\(\int dy = \int f(t)dt\)~~~~~~
\(y = F(t) + C\) where \(F\) is any anti-derivative of \(F\).
Initial Value Problem (IVP):  \(y(t_0) = y_0\)
\(y_0 = F(t_0) + C\)  implies \(C = y_0 - F(t_0)\)

Hence unique solution (if domain connected) to IVP: 
\(y = F(t) + y_0 - F(t_0)\)

 fill
1.1:  Examples of differentiable equation:

 
1.)  \(F = ma = m {dv \over dt} = mg - \gamma v\)
 

2.)  Mouse population increases at a rate proportional to the current population:

 fill

More general model :  \({dp \over dt} = rp - k\) \hfil \break
where \(p(t)\) = mouse population at time \(t\),\hfil \break
 \(r\) = growth rate or rate constant, \hfil \break
 \(k=\) predation rate = \# 
mice killed per unit time.

3.)  Continuous compounding \({dS \over dt} = rS + k\) \hfil \break
where \(S(t) =\) amount of money at time \(t\), \hfil \break
 \(r\) = interest rate, \hfil \break
 \(k=\) constant deposit rate 

direction field = slope field = graph of \({dy \over dt}\) in \(t, y\)-plane.

*** can use slope field to determine behavior of \(y\) including as $t 
\rightarrow \pm\infty$.

*** Equilibrium Solution = constant solution

Most differential equations do not 
 have an equilibrium solution.

Initial value:  A chosen point \((t_0, y_0)\) \hb
through which a solution must pass.  \hb
I.e.,  \((t_0, y_0)\) lies on the graph of \hb
 the solution that satisfies this \hb
initial value.

Initial value problem (IVP):  A differential equation where initial value is specified.

An initial value problem can have 0, 1, or multiple equilibrium solutions.

************Existence of a solution*************
 

************Uniqueness of solution*************

\end

START 100/exam1review.tex part 1

direction field = slope field = graph of \({dv \over dt}\) in \(t, v\)-plane.
 

*** can use slope field to determine behavior of \(v\) including as $t 
\rightarrow \infty$.
 

Equilibrium Solution = constant solution
 

stable, unstable, semi-stable.

{

\end

START 100/partialfrac.tex

Linear algebra pre-requisites you must know.

\({\bf b_1, ..., b_n}\) are linearly independent if 
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 
d_1{\bf b_1} + d_2{\bf b_2} + ... + d_n{\bf b_n}$ 

implies
\(c_1 = d_1\), \(c_2 = d_2\)..., \(c_n = d_n\).

or equivalently,
 
\({\bf b_1, ..., b_n}\) are linearly independent if
\u

\(c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 0\) 
implies \(c_1 = c_2 = ... c_n\).

Example 1:
\({\bf b_1} = (1, 0, 0)\), \({\bf b_2} = (0, 1, 0)\), \({\bf b_3} = (0, 0, 1)\).
 

\((1, 2, 3) \not= (1, 2, 4)\).

 
If \((a, b, c) = (1, 2, 3)\) then \(a = 1\), \(b = 2\), \(c = 3\).

  
Example 2:
\({\bf b_1} = 1\), \({\bf b_2} = t\), \({\bf b_3} = t^2\).
 
\(1 + 2t +  3t^2 \not= 1 + 2t +  4t^2\).
 

If \(a + bt +  ct^2 = 1 + 2t + 3t^2\) then \(a = 1\), \(b = 2\), \(c = 3\).

Application:  Partial Fractions

\({ 4\over (x^2 + 1)(x - 3) = {Ax + B \over x^2 + 1} + {C \over x - 3}\)}

If you don't like denominators, get rid of them:

\(\hskip .7in  {4 = (Ax + B)(x-3) + C(x^2 + 1) }\)

 
\( {4 = Ax^2 + Bx - 3Ax - 3B + Cx^2 + C\)}
 
\({4 = (A + C)x^2 + (B - 3A)x - 3B + C \)}

I.e., \(0x^2 + 0x + 4 = {(A + C)x^2 + (B - 3A)x - 3B + C }\)

Thus \(0 = A + C\),\hfill \(0 = B - 3A\), \hfill ~~\(4 = -3B + C\).  

\(C = -A\), \hfil  ~\(B = 3A\),\hfill 
 \(4 = -3(3A) + -A\) \(\Rightarrow\) \(4 = -10A\). 

Hence $A = -{2 
\over 5}\(, \)B = 3(-{2 
\over 5}) = -{6 \over 5}\(, \)C = {2 \over 5}$.
 
Thus,
${ 4\over (x^2 + 1)(x - 3) 
= {-{2 \over 5}x  - {6 \over 5} \over x^2 + 1} + {{2 \over 5} \over 
x - 3}$}

\hskip 1.65in = {${-{2}x - {6} \over 5(x^2 + 1)} + {{2} \over 
5(x - 3)}$}

 fill fill fill fill

Alternatively, can plug in \(x = 3\) to quickly find \(C\) and then solve for \(A\) and \(B\).  Can also use matrices to solve linear eqns.

\end

START 100/SPRING15/quiz1Answers.tex part 1

[10]~ 1.)  Draw the direction field for \(y' = y^2 - 1\)

\end

START 100/FALL18/quiz1_2018ans.tex part 1

\nopagenumbers

Quiz 1 ~~~SHOW ALL WORK\hfil \break
Sept 7, 2018

[10]~ 1.)  Draw the direction field for \(y' = (y + 1)(y-3)\)

\hskip 2in
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

Click on 9/10 date for solution:

\end

START 100/FALL16/e1_Fall2016ANS.tex part 1

\nopagenumbers

Math 3600 Differential Equations Exam \#1
                               Sept 30, 2016 \hfill  SHOW ALL
WORK
~~~

[10] ~1a.)  Draw the direction field for the following differential equation: 
 
\(y' = (y -2)(y + 1)^2\)
 
\includegraphics[width=41ex]{exam1figure.pdf}

[4] ~1b.)  On the direction field above, draw the solution  to the above differential equation  that satisfies the initial condition \(y(1) = 0\).

[6]~ 1c.)  Does the  differential equation whose direction field is given above have any equilibrium solutions?  If so, state whether they are stable, semi-stable or unstable.

Equilibrium solution = constant solution, \(y = c\) and thus \(y' = 0\)

\((y -2)(y + 1)^2 = 0\) implies \(y = 2, -1\)

\(y = 2\) is unstable, while \(y = -1\) is semi-stable.

[20]~ 4.)  

\includegraphics[width=48ex]{exam1fig.pdf}

4a.)  Circle the differential equation whose direction field is given above.

i.)  \(y' = y^2\)
\hfil
ii.) \(y' = (y+2)^2\)
\hfil 
iii.)  \(y' = y(y+2)^2\)

iv.) \(y' = y(y+2)\)
\hfil 
v.) 
\(y'  = {y \over y+2}\)~~~~~~~~~~~~~
\hfil 
vi.) 
\(y'  = {y +2 \over y}\)~~~~~~~~~~~~~

vii.) \(y'  = {y^{2} \over t}\)
\hfil 
viii.) \(y'  = t qrt{y} \)~~~~~~~
\hfil 
ix.) 
\(y'  = {t  y^{2}}\)~~~~~~~~~

4b.)  Draw the solution to the differential equation  whose direction field is given above that satisfies the initial condition \(y(1) = -3\).

4c.)  Does the  differential equation whose direction field is given above have any equilibrium solutions?  If so, state whether they are stable, semi-stable or unstable.

\end

START 100/FALL18/e1_Fall2018ansOLD.tex part 1

\nopagenumbers

Math 3600 Differential Equations Exam \#1
                               Sept 26, 2018 \hfill  SHOW ALL
WORK
~~~

[10] ~1a.)  Draw the direction field for the following differential equation: 
{\(y' = y(2-y )\)}
\includegraphics[width=30ex]{grapht-eps-converted-to.pdf}
[4] ~1b.)  On the direction field above, draw the solution  to the above differential equation  that satisfies the initial condition \(y(0) = 1\).

~~~~~Note the solution satisfying the initial condition \(y(0) = 1\) must pass thru the initial value (0, 1).

[6]~ 1c.)  Does the  differential equation whose direction field is given above have any equilibrium solutions?  If so, state whether they are stable, semi-stable or unstable.

~~~~~An equilibrium solution is a constant solution, \(y = C\).  
~~~~~The graph of a constant solution, \(y= C\) is a horizontal line.
~~~~~\(y = 2\) is stable and
\(y = 0\) is unstable

\end

START 100/SPRING13/e1_2013ANS.txt part 1

\nopagenumbers

Math 100 Differential Equations Exam \#1
                               February 27, 2013 \hfill  SHOW ALL
WORK
~~~

[10]~ 1.) By giving a specific example, prove that \(f: R \rightarrow R, ~f(x) = e^x\) is not onto.

Answer 1:  We know that \(e^x > 0\) for all \(x\).  Thus there is no \(x\) such that \(f(x) = e^x = -1\).  Hence -1 is not in the image of \(f\).

Answer 2:  Suppose \(f(x) = -1\).  Then \(e^x = -1\).  But then \(x = ln(-1)\).  But \(ln(-1)\) is not defined.  Thus -1 is not in the image of \(f\).

[15]~ 3.)  Draw the direction field for \(y' = {1 \over 2}y + 1\).  Determine if there are any equilibrium solutions.  If so,  determine if the equilibrium solution(s) are stable, unstable or semi-stable.

Equilibrium solution = constant solution.  Thus \(y' = 0\).  

\( {1 \over 2}y + 1 = 0\) implies \( {1 \over 2}y = -1\).  Thus \(y = -2\) is the equilibrium solution.  

\end

START 34/FALL03/finalexamANS.txt part 1

\nopagenumbers

Math 34 Differential Equations Final Exam
                               December 15, 2003 \hfill  SHOW ALL
WORK
~

[10]~ 7.)  Draw the direction field for the differential equation \(y' = (y-2)(y + 2)\).  
Based on
the direction field, determine the behavior of \(y\) as \(t \rightarrow \infty\).  If this behavior 
depends on the initial value of \(y\) at \(t=0\), describe this dependency.

If \(y(0) > 2\), then \(lim_{t \rightarrow \infty} y(t) = +\infty\)

If \(y(0) = 2\) or \(-2\), then \(lim_{t \rightarrow \infty} y(t) = 2\)

If \( y(0) < 2\), then \(lim_{t \rightarrow \infty} y(t) = -2\)

\end

START 34/FALL03/finalreviewANS2.txt part 1

Note the following review problems DO NOT cover all problem types which may appear on the 
final.  

11.)  For each of the following, draw the direction field for the given differential equation.  
Based on
the direction field, determine the behavior of \(y\) as \(t \rightarrow \infty\).  If this behavior 
depends on the initial value of \(y\) at \(t=o\), describe this dependency.

11a.)  \(y' = y\)

11a.)  \(y' = 1\)

11a.)  \(y' = y(y + 4)\)

\end

START 100/SPRING15/e1_2016ans.tex part 1

[20]~ 4.)  

4a.)  Circle the differential equation whose direction field is given above.

\hfil 
v.) 
\(y'  = {y \over y+2}\)~~~~~~~~~~~~~
\hfil 

4b.)  Draw the solution to the differential equation  whose direction field is given above that satisfies the initial condition \(y(1) = -3\)

4c.)  Does the  differential equation whose direction field is given above have any equilibrium solutions?  If so, state whether they are stable, semi-stable or unstable.

The constant solution \(y = 0\) is an unstable equilibrium solution.

\end

START 100/SPRING15/quiz1Answers.tex part 1

[10]~ 2.)  Solve \(sin(t) y' + y cos(t) = 5\) 
 \hfil \break(hint:  this is a short problem if you are observant).

\(sin(t) y' + y cos(t) = 5\)

 \((sin(t) y)'  = 5\)  
 
  \(\int (sin(t) y)' dt  = \int 5 dt\)  

 \( in(t) y = 5t + C\)  

 \( y  = {5t + C \over sin(t)}\)  

START 100/FALL17/quiz1Fall2017Bans.tex part 1

[14]~ 2.)  Solve and state where the solution is defined:  \(t^5y' + 5t^4y = {1 \over t}\)

\(t^5y' + 5t^4y = {1 \over t}\)

\((t^5y)' = {1 \over t}\)

\(\int(t^5y)'dt = \int{1 \over t}dt\)

\(t^5y = ln|t|+C\)

\(y = t^{-5}(ln|t| + c)\)

Solution:  \(\underline{y = t^{-5}(ln|t| + c)}\) 
\hfill
Domain:  \(\underline{t \not= 0 \hbox{ or equivalently } (-\infty, 0)\cup (0, \infty)}\) 

\end{document}

START 100/FALL17/quiz1Fall2017ans.tex part 1

[14]~ 2.)  Solve and state where the solution is defined:  \(t^4y' + 4t^3y = {1 \over t}\)

\(t^4y' + 4t^3y = {1 \over t}\)

\((t^4y)' = {1 \over t}\)

\(\int(t^4y)'dt = \int{1 \over t}dt\)

\(t^4y = ln|t|+C\)

\(y = t^{-4}(ln|t| + c)\)

Solution:  \(\underline{y = t^{-4}(ln|t| + c)}\) 
\hfill
Domain:  \(\underline{t \not= 0 \hbox{ or equivalently } (-\infty, 0)\cup (0, \infty)}\) 

\end

START 100/ch1and2a.tex part 2

1.3:

ODE (ordinary differential equation):  single independent variable

~~~Ex:  \({dy \over dt} = ay + b\)

PDE (partial differential equation):  several independent variables

~~~Ex:  \({\partial xy \over \partial x} = {\partial xy \over \partial y}\)

order of differential eq'n:  order of highest derivative
 

example of order \(n\):  \(y^{(n)} = f(t, y, ..., y^{(n-1)})\)

Linear vs Non-linear
 
Linear:  \(a_0 y^{(n)} + ... + a_{n-1}y' + a_ny = g(t)\) 

where \(a_i\)'s are functions of \(t\)

Note for this linear equation, the left hand side is a linear combination of the derivatives of \(y\) (denoted by \(y^{(k)}, k = 0, ..., n\))  where the coefficient of  \(y^{(k)}\) is a function of \(t\) (denoted \(a_k(t)\)).

Linear:  \(a_0(t) y^{(n)} + ... + a_{n-1}(t)y' + a_n(t)y = g(t)\)

Determine if linear or non-linear:
 
Ex:  \(t y'' - t^3y' - 3y = sin(t)\)
 
Ex:  \( 2y'' - 3y' - 3y^2 = 0\)

Show that for some value of \(r\), \(y = e^{rt}\) is a soln to the \(1^{rst}\) order linear homogeneous equation \(2y' - 6y = 0\).
To show something is a solution, plug it in:
~~~\(y = e^{rt}\) implies \(y' = re^{rt}\). \hfill Plug into 
\(2y' - 6y = 0\):
~~~\(2re^{rt} + 6e^{rt} = 0\) implies \(2r - 6 = 0\) implies \(r = {3}\)
Thus \(y = e^{3t}\) is a solution to \(2y' - 6y = 0\).
Show  \(y = Ce^{3t}\) is a solution to \(2y' - 6y = 0\).
$2y' - 6y = 2(Ce^{3t})' - 6(Ce^{3t}) = 
2C(e^{3t})' - 6C(e^{3t})$

 \rightline{ = 
$C[2(e^{3t})' - 6(e^{3t})] = C(0) 
=0$.}
If \(y(0) = 4\), then \(4 = Ce^{3(0)} \) implies \(C = 4\).
 Thus by existence and uniqueness thm, \(y = 4e^{3t}\) is the unique solution to IVP: \(2y' + 6y = 0\),  \(y(0) = 4\).

\end

START 34/FALL10/quiz1ANS.tex

SHOW ALL WORK
~~~

 1.)  Give an example of a linear differential equation:

~~~\(y' + y = 0\)  or \(sin(t)y y' + t^{1 \over 2} y = ln(t)/e^t\) or ...

 2.)  Give an example of a non-linear differential equation:

~~~\(y' y = 0 \) or ....

3.)  From section 1.4, name one mathematician who studied differential equations OR one application (but not related to gravity) of
differential equations.

~~~Newton or mechanics or ...

4.)  Circle T for True or F for False:

4a.) Numerical approximations for solutions to differential equations are often needed as the solutions 
to most differential equations cannot be expressed algebraically.

4b.)  If a computer is used to find a numerical approximation to a differential 
equation, then we know the equation has at least one solution.

\end

START 100/FALL17/quiz1Fall2017ans.tex part 2

\nopagenumbers

Quiz 1 Form {\bf A}\hfil \break
Sept 1, 2017

1.) Determine the \underbar{order} of the given differential equations and also state whether the equation is  \underbar{linear or nonlinear}:

[3]~ 1a.)  \(t + yy' = 1\) is a 
\(\underline{first}\) order \(\underline{nonlinear}\) differential equation.

[3]~ 1b.)  \(ty + y' = 1\) is a 
\(\underline{first}\) order \(\underline{linear}\) differential equation.

\end{document}

\end 

START 100/FALL17/quiz1Fall2017Bans.tex part 2

\nopagenumbers

Quiz 1 Form {\bf B}\hfil \break
Sept 1, 2017

1.) Determine the \underbar{order} of the given differential equations and also state whether the equation is  \underbar{linear or nonlinear}:

[3]~ 1a.)  \(ty + y' = 1\) is a 
\(\underline{first}\) order \(\underline{linear}\) differential equation.

[3]~ 1b.)  \(t + yy' = 1\) is a 
\(\underline{first}\) order \(\underline{nonlinear}\) differential equation.

\end 

Answer:\(\underline{~~ y  = {5t + C \over sin(t)}~~}\)
\end 

START 100/quiz2Fall2019Empty.tex part 1

\nopagenumbers

Quiz 2 Section 9*     ~~~~~~Sept 27, 2019

[5]~ 1.)  Solve  the following first order linear differential equation:

 fill

General Solution:  \(\underline{\hskip 3.3in}\) 

We will cover the following next week

1.) If \(\phi\) is a solution to a first order linear differential equation, then \(c\phi\) is also a 
solution to this equation.
\hfill   A)  True  \hfill  B)False \hfill ~~

2.) If \(\phi\) is a solution to a first order linear homogeneous differential equation, then \(c\phi\) is 
also a 
solution to this equation.

\hfill   A)  True  \hfill  B)False \hfill ~~

3.) If \(\phi\) is a solution to a first order linear homogeneous differential equation with constant 
coefficients, then \(c\phi\) is also a 
solution to this equation.

\hfill   A)  True  \hfill  B)False \hfill ~~

---

\end

START 100/FALL18/quiz2_2018ans.tex part 1

Quiz 2 ~~~SHOW ALL WORK\hfil \break
Sept 20, 2018

[10]~ 1.) Solve \(ty' + (t+1) y = t\) \hfill (Compare to 2.1: 12)

\(y' + {(t+1)\over t} y = 1\)

$u(t) = e^{\int  {(t+1)\over t} dt } = e^{\int  (1+{1\over t}) dt } = 
e^{t+ln| t|} = e^te^{ln| t|} = |t|e^{t}$.

Let \(u(t) = te^t\)

\( te^ty' + { e^t(t+1)} y =  te^t\)

\( (te^ty)' =  te^t\) \hfill {\bf Check:}  \((te^ty)' =   te^ty' +  (e^t + te^t)y \) \hfill~\hfill~

\( \int (te^ty)' =  \int te^t\) 

 integration by parts:

~~~~~\(u = t~~~~~~~~~dv = e^t\)  skip -9pt
~~~~~\(du = 1~~~~~~~v =e^t\)  skip -9pt
~~~~~\(d^2u = 0~~~~~\int v =e^t\)

\( te^ty =  te^t - e^t + C\)  

Thus \(y =  1 - t^{-1} + Ct^{-1}e^{-t}\)

\hskip 1in
Answer:\(\underline{~~y =  1 - t^{-1}e^t + Ct^{-1}e^{-t}~~}\) 

START 100/FALL18/e1_Fall2018ansOLD.tex part 2

Math 3600 Differential Equations Exam \#1
                               Sept 26, 2018 \hfill  SHOW ALL
WORK
~~~

[20]~ 4.) Find the general solution to  \(ty' - 2y= t^3 e^t - 8\).  Also find the solution that passes thru the point (1, 3). 
How does the solution passing thru (1, 3)
behave as \(t \rightarrow \infty\)?

 \(ty' - 2y= t^3 e^t - 8\) implies  \(1y' + (-{2 \over t})y= t^2 e^t - 8 t^{-1}\). 

Integrating factor:  \(u = e^{\int p(t)dt} = e^{\int -{2 \over t}dt } = e^{-2ln|t|} = e^{ln(|t|^{-2})} = t^{-2}\)

Let \(u(t) = t^{-2}\)

 \(t^{-2}y'  -{2  t^{-3}}y=  e^t - {8 t^{-3}}\)

 \((t^{-2}y)'  =  e^t - {8 t^{-3}}\) \hfill {\bf Check: \((t^{-2}y)'  = t^{-2}y'  -{2  t^{-3}}y \) }

 \(\int (t^{-2}y)'dt  = \int( e^t - {8 t^{-3}})dt\)

 \(t^{-2}y  =  e^t + {4 t^{-2}}+C\)

 \(y  =  t^2e^t + {4}+Ct^2\)

\(y(1) = 3\):  ~~ \(3  =  (1)^2e^1 + {4}+C(1)^2 = e + 4 + C\) 

\(3  = e + 4 + C\).  Thus \(C = -1-e\)

IVP soln:   
 \(y  =  t^2e^t + {4}+(-1-e)t^2\)

 \(y  =  t^2e^t + {4}-t^2-et^2\)

 \(y  =  t^2(e^t - e - 1) + {4}\)

 \(t \rightarrow +\infty\),  \(t^2 \rightarrow +\infty\) and  \(e^t\rightarrow +\infty\) and hence 
  \(e^t - e - 1\rightarrow +\infty\) .

Thus  \(y  =  t^2(e^t - e - 1) + {4}\rightarrow +\infty\) 

 fill
General solution: \(\underline{~~y  =  t^2e^t + {4+Ct^2~~}\)}
IVP solution: \(\underline{~~ y  =  t^2(e^t - e - 1) + {4~~}\)}
 \(t \rightarrow \infty\),  \(y \rightarrow \underline{~~+\infty~~\)}

\end

START 100/ch1and2a.tex part 3

CH 2:  Solve \({dy \over dt} = f(t, y)\) for special cases:
2.2:  Separation of variables: \(N(y)dy = P(t)dt\)
2.1:  First order linear eqn:  \({dy \over dt} + p(t)y = g(t)\)

Ex 1:  \(t^2y' + 2ty = t sin(t)\)
 
Ex 2:  \(y' = ay + b\)
 
Ex 3:  \(y' + 3t^2y = t^2\), \(y(0) = 0\)
Note:  can use either section 2.1 method (integrating factor) or 2.2 method (separation of variables) to solve ex 2 and 3.

Ex 1:  \(t^2y' + 2ty = sin(t)\) \hfil \break
(note, cannot use separation of variables).
\(t^2y' + 2ty = sin(t)\)
~~~\( (t^2y)' = sin(t)\)
~~~implies~~~
\( \int (t^2y)' dt = \int sin(t) dt\)
\( (t^2y) = -cos(t) + C\)
implies
\( y = -t^{-2}cos(t) + Ct^{-2}\)

Ex. 2:  Solve \({dy \over dt} = ay + b\) by separating variables:

\({dy \over ay + b} = dt\)
\(~~\Rightarrow~~\)
\(\int {dy \over ay + b} = \int dt\)
\(~~\Rightarrow~~\)
\({ln|ay + b| \over a} = t + C\)

\({ln|ay + b|} = at + C\)
~~~~~implies~~~~~
\(e^{ln|ay + b|} = e^{at + C}\)

\(|ay + b| = e^C e^{at}\)
~~~~~implies~~~~~
\(ay + b = \pm(e^C e^{at})\)

\(ay  = C e^{at} - b\)
~~~~~implies~~~~~
\(y  = C e^{at} - {b \over a}\)

Gen ex:  Solve \(y' + p(x)y = g(x)\)
Let \(F(x)\) be an anti-derivative of \(p(x)\). %\hfil \break
Thus \(p(x) = F'(x)\)
 
\(e^{F(x)} y' + [p(x) e^{ F(x)} ] y = g(x)e^{F(x)}\)
 
\(e^{F(x)} y' + [F'(x) e^{ F(x)} ] y = g(x)e^{F(x)}\)
 
~~~~~~\([e^{F(x)} y]' ~~= ~~g(x)e^{F(x)}\)
 

\(e^{F(x)} y = \int g(x)e^{F(x)} dx\)
 
\( y = e^{-F(x)} \int g(x)e^{F(x)} dx\)

\end

START 100/exam1review.tex part 2

{\bf Solving first order differential equation:}

Method 1 (sect. 2.2):  Separate variables.

Method 2 (sect. 2.1):  If linear [\(y'(t) + p(t)y(t) = g(t)\)], multiply equation by 
an integrating factor \hfil \break $u(t) = 
e^{\int p(t)dt}$.  
\(y' + py = g\)
\(y'u + upy = ug\)

\( (uy)' = ug\)

\( \int(uy)' = \int ug\)

\( uy = \int ug\)

etc...

\end

START 100/FALL18/quiz1_2018ans.tex part 2

\nopagenumbers

Quiz 1 ~~~SHOW ALL WORK\hfil \break
Sept 7, 2018

[10]~ 2.)  Solve \(y' = {1 \over x(2y+3)}\), ~\(y(1) =-2 \)

\({dy \over dx} = {1 \over x(2y+3)}\) \hskip 2in Compare to 2.2:  14

Separate variables:  \((2y+3)dy = {dx \over x}\)

\(y^2 + 3y = ln|x| + C\).

\(y^2 + 3y - ln|x| - C = 0\).

\(y = {-3 \pm  qrt{9 - 4 (- ln|x| - C)} \over 2}\)
\( = {-3 \pm  qrt{9 + 4 ln|x| + C} \over 2}\)
\( = {-3 \pm  qrt{4 ln|x| + C} \over 2}\)

General solution:  \( y= {-3 \pm 2 qrt{ ln|x| + C} \over 2}\)

IVP:  ~\(y(1) =-2 \)

\(-2 =  {-3 \pm 2 qrt{ ln|1| + C} \over 2}=  {-3 \pm 2 qrt{ C} \over 2}\)

\(-4 = {-3 \pm 2 qrt{ C} }\).  Thus \(-1 = { \pm 2 qrt{ C} }\).  Hence   \(-1 = { - 2 qrt{ C} }\).

Thus \( qrt{ C} = {1 \over 2}\) and \(C = {1 \over 4}\)

Hence IVP solution:     $ y= {-3 - 2 qrt{ ln|x| + {1 \over 4}} \over 2}
y= {-3 - 2 qrt{ {4ln|x| + 1 \over 4}} \over 2}$
\( = {-3 -  qrt{4 ln|x|  + 1} \over 2}\)

Answer:\(\underline{~~y= {-3 -  qrt{1 + 4 ln|x| } \over 2}~~}\)

\end

START 100/FALL18/quiz2_2018ans.tex part 2 & 100/quiz4_2018ans.tex part 1

\({dy \over dx} = {1 \over x(2y+3)}\)

Separate variables:  \((2y+3)dy = {dx \over x}\)

\(y^2 + 3y = ln|x| + C\).

\(y^2 + 3y - ln|x| - C = 0\).

\(y = {-3 \pm  qrt{9 - 4 (- ln|x| - C)} \over 2}\)

\(y = {-3 \pm  qrt{9 + 4 ln|x| + C} \over 2}\)

\(y = {-3 \pm  qrt{9  + C)} \over 2}\)

\(y = {-3 \pm  qrt{9 -8)} \over 2}\)

\(y = {-3 - 1 \over 2}\)

\(y = {-3 \pm  qrt{9 + 4 ln|x| -8} \over 2}\)

\(y = {-3 -  qrt{1 + 4 ln|x| } \over 2}\)

2.2       14

\end 

START 100/FALL16/e1_Fall2016ANS.tex part 2

[15]~ 2.)  Solve the initial value problem {\bf{ for \(y\)}}: ~~ \(y' + {3x \over y - 4} = 0 \), \(y(1) = -2\).

 \({dy \over dx}   = -{3x \over y - 4}\)

\(\int(y-4)dy = \int -3xdx\)

\({y^2 \over 2} - 4y = -{3 \over 2}x^2\)

\(y^2 - 8y = -3x^2 + C\)

\(y^2 - 8y + 3x^2 + C = 0\)

\(y = {8 \pm   qrt{64 - 4(3x^2 + C)} \over 2} = 4 \pm  qrt{16 - 3x^2 + C} = y\)

 \(y(1) = -2\):  \(-2 =  4 \pm  qrt{16 - 3(1)^2 + C}\) implies \(-6 =   - qrt{16 - 3 + C}\)

Note initial value determines sign of \(\pm\).   In this case, IVP only has a solution when we choose the negative sign.  The the IVP in this case means \(y = 4 -  qrt{16 - 3x^2 + C}\) where we determine \(C\) below:

\(36 =  {13 + C}\).  Thus \(C = 36 - 13 = 23\) and \(y =  4 -  qrt{16 - 3x^2 + 23} =  4 -  qrt{39 - 3x^2}\)

 fill
Answer: \(\underline{~~y =  4 -  qrt{39 - 3x^2~~}\)}

\end

START 100/SPRING13/final3600ANS.tex part 1

22M:100 (MATH:3600:0001) Final Exam \hfil \break
May 15, 2013 \hfill SHOW ALL STEPS \hfill

[13]~ 1.)   Solve:  \(ty' + 6t^2y - t^2 = 0\), \(y(1) = 3\)

\(y' + 6ty  = t\), \(\int 6tdt = 3t^2\)

\(e^{3t^2}y' + 6te^{3t^2} =  te^{3t^2}\)
~~implies~~
\([e^{3t^2}y]' = te^{3t^2}\)
~~implies~~
\(\int [e^{3t^2}y]'dt = \int te^{3t^2}dt\)

implies
\(e^{3t^2}y = \frac{e^{3t^2}}{6} + C\) (via \(u\)-substitution).~~~
Thus \(y = \frac{1}{6} +  Ce^{-3t^2}\)

\(y(1) = 3\):~~~~  \(3 = \frac{1}{6} + Ce^{-3}\)
implies \(C = \frac{17}{6}e^3\)
 fill
Answer: \underline{~~\(y = \frac{1{6} +  \frac{17}{6}e^{-3t^2+3}\)~~}}

\end

START 34/FALL03/finalexamANS.txt part 2

\nopagenumbers

Math 34 Differential Equations Final Exam
                               December 15, 2003 \hfill  SHOW ALL
WORK
~

[10]~ 6.) Solve the following initial value problem:  \({y' \over t^2} = -3y + 1, ~y(0) = 0\)

Method 1:  separation of variables

\({dy \over dt} = t^2(-3y + 1)\)

\(\int{dy \over -3y + 1} = \int t^2dt\)

\({1 \over -3}ln |-3y + 1| = {1 \over 3} t^3 + C\)

\(ln |-3y + 1| = -t^3 + C\)

\(|-3y + 1| = e^{-t^3} e^C\)

\(-3y  = Ce^{-t^3} - 1\)

\(y  = Ce^{-t^3} + {1 \over 3}\)

Method 2:  Integrating Factor

\(y' + 3t^2y = t^2\).     ~~~Let \(u = e^{\int 3t^2 dt} = e^{t^3}\)

\(e^{t^3}y' + 3t^2e^{t^3}y = t^2e^{t^3}\)

\((e^{t^3}y)'  = t^2e^{t^3}\)

\(\int (e^{t^3}y)'  = \int t^2e^{t^3} dt\). Integration by substitution:  Let \(u = t^3\), $du = 
3t^2dt$

\(e^{t^3}y = {1 \over 3}\int e^u du\)

\(e^{t^3}y = {1 \over 3} e^u  + C = {1 \over 3} e^{t^3} + C\)

\(y = {1 \over 3}  + Ce^{-t^3}\)

\(y(0) = 0\): 0 = \({1 \over 3}  + C\).  Hence \(C = -{1 \over 3} \)
 fill
 fill
Answer 6.) \(\underline{y  = -{1 \over 3 e^{-t^3} + {1 \over 3}}\)}

\end

START 100/SPRING15/quiz2_3600ans.tex part 1

\nopagenumbers

Quiz 2 \hfill  Show your work \break
Feb 19, 2016 \hfill  Circle your answer.

[10]~ 2.)  Solve: \(y' = y~sin(x) + y\).

Separate variables:  \({dy \over dx} = y(sinx + 1)\).

\({dy \over y} = (sinx + 1)dx\)

\(\int {dy \over y} = \int (sinx + 1)dx\)

\(ln|y| = -cos(x) + x + C\)

\(ln|y| = -cos(x) + x + C\)

\(|y| = e^{-cos(x) + x + C} = e^Ce^{x-cos(x)} \)

Thus \(y = Ce^{x-cos(x)}\)

Answer:\(\underline{~~y = Ce^{x-cos(x)~~}\)}
\end 

START 100/SPRING15/quiz2answers.tex part 1

\nopagenumbers

Quiz 2 \hfill  Show your work \break
Feb 19, 2016 \hfill  Circle your answer.

[10]~ 2.)  Solve: \(y' = ycos(x)\).

Separate variables:  \({dy \over dx} = ycos(x)\).

\({dy \over y} = cos(x)dx\)

\(\int {dy \over y} = \int cos(x)dx\)

\(ln|y| = sin(x) + C\)

\(ln|y| = sin(x) +  C\)

\(|y| = e^{sin(x) + C} = e^Ce^{sin(x)} \)

Thus \(y = Ce^{sin(x)}\)

Answer:\(\underline{~~y = Ce^{sin(x)~~}\)}

 fill
 fill
 fill
 fill

\end 

START 100/2_3ex.tex

2.3: Modeling with differential equations.
Suppose salty water enters and leaves a tank at a rate of 2 liters/minute.
  Suppose also that the salt concentration of the water entering the tank varies with respect to time according to \(Q(t) \cdot t sin(t^2)\) g/liters where \(Q(t) = \) amount of salt in tank in grams. (Note:  this is not realistic).
  \t
    If the tank contains 4 liters of water and initially contains 5g of salt, find a formula for the amount of salt in the tank after \(t\) minutes.

Let \(Q(t) = \) amount of salt in tank in grams.

Note \(Q(0) = 5\) g

rate in = (2 liters/min)(\(Q(t) \cdot t sin(t^2)\) g/liters) 

 rate out =   (2 liters/min)(\({Q(t) g \over 4 \hbox{liters}}\)) = \({Q \over 2}\) g/min
 

\({dQ \over dt}\) = rate in - rate out = \(2Qtsin(t^2) - {Q \over 2}\)

\({dQ \over dt} = Q(2tsin(t^2) - {1 \over 2})\)

This is a first order linear ODE.  It is also a separable ODE.  Thus can use either 2.1 or 2.2 methods.
Using the easier 2.2:

\(\int {dQ \over Q} = \int (2tsin(t^2) - {1 \over 2})dt = \int 2tsin(t^2)dt - \int{1 \over 2}dt\)

Let \(u = t^2\), \(du = 2tdt\)

\(ln|Q| = \int sin(u)du - {t \over 2}\)  = \(-cos(u) - {t \over 2} + C\)
  skip 5pt
 \rightline{= \(-cos(t^2) - {t \over 2} + C\)}

\(|Q| = e^{-cos(t^2) - {t \over 2} + C} = e^C e^{-cos(t^2) - {t \over 2}}\)

\(Q = C e^{-cos(t^2) - {t \over 2}}\)

\(Q(0) = 5:  ~~~ 5 = Ce^{-1 - 0} = Ce^{-1}\).  Thus \(C = 5e\)

Thus \(Q(t) =  5e \cdot e^{-cos(t^2) - {t \over 2}}\)
Thus \(Q(t) =  5  e^{-cos(t^2) - {t \over 2 + 1}\)}

Long-term behaviour:

\(Q(t) =  5 ( e^{-cos(t^2)} )(e^{-t \over 2})e\)

As \(t \rightarrow \infty\), \(e^{-t \over 2} \rightarrow 0\), while \(5 ( e^{-cos(t^2)} )e\) are finite.  

Thus
as \(t \rightarrow \infty\), \(Q(t)  \rightarrow 0\).

\end

START 100/ch1and2a.tex part 4

2.3: Modeling with differential equations.
Suppose salty water enters and leaves a tank at a rate of 2 liters/minute.
  Suppose also that the salt concentration of the water entering the tank varies with respect to time according to \(Q(t) \cdot t sin(t^2)\) g/liters where \(Q(t) = \) amount of salt in tank in grams. (Note:  this is not realistic).
  \t
    If the tank contains 4 liters of water and initially contains 5g of salt, find a formula for the amount of salt in the tank after \(t\) minutes.

Let \(Q(t) = \) amount of salt in tank in grams.

Note \(Q(0) = 5\) g

rate in = (2 liters/min)(\(Q(t) \cdot t sin(t^2)\) g/liters) 

 rate out =   (2 liters/min)(\({Q(t) g \over 4 \hbox{liters}}\)) = \({Q \over 2}\) g/min
 

\({dQ \over dt}\) = rate in - rate out = \(2Qtsin(t^2) - {Q \over 2}\)

\({dQ \over dt} = Q(2tsin(t^2) - {1 \over 2})\), ~~~ \(Q(0) = 5\)

This is a first order linear ODE.  It is also a separable ODE.  Thus can use either 2.1 or 2.2 methods.
 
Using the easier 2.2:

\(\int {dQ \over Q} = \int (2tsin(t^2) - {1 \over 2})dt = \int 2tsin(t^2)dt - \int{1 \over 2}dt\)
Let \(u = t^2\), \(du = 2tdt\)

\(ln|Q| = \int sin(u)du - {t \over 2}\)  = \(-cos(u) - {t \over 2} + C\)
  skip 5pt
 \rightline{= \(-cos(t^2) - {t \over 2} + C\)}

 
\(|Q| = e^{-cos(t^2) - {t \over 2} + C} = e^C e^{-cos(t^2) - {t \over 2}}\)

\(Q = C e^{-cos(t^2) - {t \over 2}\)}

\(Q(0) = 5:  ~~~ 5 = Ce^{-1 - 0} = Ce^{-1}\).  Thus \(C = 5e\)

Thus \(Q(t) =  5e \cdot e^{-cos(t^2) - {t \over 2}}\)
Thus \(Q(t) =  5  e^{-cos(t^2) - {t \over 2 + 1}\)}
Long-term behaviour:

\(Q(t) =  5 ( e^{-cos(t^2) )(e^{-t \over 2})e\)}
As \(t \rightarrow \infty\), \(e^{-t \over 2} \rightarrow 0\), while \(5 ( e^{-cos(t^2)} )e\) are finite.  

Thus
as \(t \rightarrow \infty\), \(Q(t)  \rightarrow 0\).

\end

START 100/quiz2Fall2019Empty.tex part 2

3.) Word problem:  State the initial value problem describing ...  Do {\bf NOT} solve.

 fill fill fill fill fill fill fill fill fill fill fill fill fill fill fill
 fill
[4]~~ Differential equation:  \(\underline{\hskip 3in}\) 

[2]~~ Initial Value:  \(\underline{\hskip 3in}\) 

[3]~ 4.)  The solution to the initial value problem \(y' = {1 - x \over y}\), \(y(0) = - qrt{3}\) is\hb  \(y = -  qrt{-x^2 + 2x + 3}\).~~
State the largest interval on which the solution is defined.

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 + 2y - 8\), \(y(0) = 3\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 Note \(y(0) = 3 > 2\).  Thus 
\(y \rightarrow +\infty\) as \(t \rightarrow +\infty\) 

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 + 2y - 8\), \(y(0) = 1\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 Note \(y(0) = 1 \in (-4, 2)\).  Thus 
\(y \rightarrow -4\) as \(t \rightarrow +\infty\) 

---

\end

START 100/2_3ex.tex

2.3: Modeling with differential equations.
Suppose salty water enters and leaves a tank at a rate of 2 liters/minute.
  Suppose also that the salt concentration of the water entering the tank varies with respect to time according to \(Q(t) \cdot t sin(t^2)\) g/liters where \(Q(t) = \) amount of salt in tank in grams. (Note:  this is not realistic).
  \t
    If the tank contains 4 liters of water and initially contains 5g of salt, find a formula for the amount of salt in the tank after \(t\) minutes.

Let \(Q(t) = \) amount of salt in tank in grams.

Note \(Q(0) = 5\) g

rate in = (2 liters/min)(\(Q(t) \cdot t sin(t^2)\) g/liters) 

 rate out =   (2 liters/min)(\({Q(t) g \over 4 \hbox{liters}}\)) = \({Q \over 2}\) g/min
 

\({dQ \over dt}\) = rate in - rate out = \(2Qtsin(t^2) - {Q \over 2}\)

\({dQ \over dt} = Q(2tsin(t^2) - {1 \over 2})\)

This is a first order linear ODE.  It is also a separable ODE.  Thus can use either 2.1 or 2.2 methods.
Using the easier 2.2:

\(\int {dQ \over Q} = \int (2tsin(t^2) - {1 \over 2})dt = \int 2tsin(t^2)dt - \int{1 \over 2}dt\)

Let \(u = t^2\), \(du = 2tdt\)

\(ln|Q| = \int sin(u)du - {t \over 2}\)  = \(-cos(u) - {t \over 2} + C\)
  skip 5pt
 \rightline{= \(-cos(t^2) - {t \over 2} + C\)}

\(|Q| = e^{-cos(t^2) - {t \over 2} + C} = e^C e^{-cos(t^2) - {t \over 2}}\)

\(Q = C e^{-cos(t^2) - {t \over 2}}\)

\(Q(0) = 5:  ~~~ 5 = Ce^{-1 - 0} = Ce^{-1}\).  Thus \(C = 5e\)

Thus \(Q(t) =  5e \cdot e^{-cos(t^2) - {t \over 2}}\)
Thus \(Q(t) =  5  e^{-cos(t^2) - {t \over 2 + 1}\)}

Long-term behaviour:

\(Q(t) =  5 ( e^{-cos(t^2)} )(e^{-t \over 2})e\)

As \(t \rightarrow \infty\), \(e^{-t \over 2} \rightarrow 0\), while \(5 ( e^{-cos(t^2)} )e\) are finite.  

Thus
as \(t \rightarrow \infty\), \(Q(t)  \rightarrow 0\).

\end

START 100/FALL17/quiz2Fall2017ans.tex

\nopagenumbers

Quiz 2 Form A \hfil \break
Sept 15, 2016

[10] 1i.  Suppose \\(75 is invested at an annual rate of return \)r$ compounded continuously.
State the initial value problem describing the amount of money after \(t\) years.

Differential equation:  \(\underline{~~y' = ry~~}\) 

Initial Value:  \(\underline{~~y(0) = 75~~}\) 

1ii. Circle the general solution to the differential equation in problem 1:

D.)  \(y = Ce^{rt}\) \hfill 

1iii.  Circle the solution to the initial value problem in problem 1:

D.)  \(y = 75e^{rt}\) \hfill 

[10]~ 2.)  Suppose water containing 3 lbs of salt per gallon  enters and leaves a tank at a rate of 8 gallons/hour.  Suppose  the tank originally contains 7 lbs of salt in 500 gallons of water.  State the initial value problem describing the amount of salt in the tank at time \(t\).  Do NOT solve.

Differential equation:  \(\underline{~~Q' = 24 - {8Q \over 500}~~ or~~ Q' = 24 - {2Q \over 125}}\) 

Initial Value:  \(\underline{~~Q(0) = 7~~}\) 

Let \(Q(t) =\) amount of salt (in pounds) in the tanke at time \(t\).

\({dQ \over dt} =\) Rate in - rate out \(= ({3lbs \over 1 gallon})({8 gallons \over 1 hour}) -  ({Q(t)lbs \over 500 gallon})({8 gallons \over 1 hour})\)

\(Q' = 24 - {8Q \over 500} =  24 - {4Q \over 250} = 24 - {2Q \over 125}\)

\end{document}

START 34/FALL03/finalexamANS.txt part 3

[10]~ 11.)  Suppose that a sum \(S_0\) is invested at an annual rate of return \(r\) compounded 
continuously.  Find the time \(T\) required for the original sum to double in value as a 
function of \(r\). Assume that the rate of change of the value of the investment is equal to 
the interest rate \(r\) times the current value of the investment \(S(t)\).

\({dS \over dt} = rS(t)\)~~~~~~~~~
\({dS \over S(t)} = rdt\)~~~~~~~~~
\(ln|S| = {rt} + C\)~~~~~~~~~~~
\(|S| = e^{rt}e^C\)~~~~~~~~~~~~

\(S = Ce^{rt}\)

\(S(0) = S_0\):  \(S_0 = Ce^{0} = C\)

\(S = S_0e^{rt}\)

\(2S_0 = S_0e^{rt}\),~~~~~~~~~
\(2 = e^{rt}\),~~~~~~~~~~
\(ln(2) = {rt}\)

\(t = {ln(2) \over r}\)

Answer 11.) \(\underline{t = {ln(2) \over r}\)}

\end

START 100/exam1review.tex part 3

Method 3 (sect. 2.4):  Solve Bernoulli's equation, 
\[y' + p(t)y = g(t)y^n,\] 

when \(n > 1\) by changing it
to a linear equation by
substituting \(v = y^{1-n}\)

Section 2.4:  Existence and Uniqueness.

{\bf In general,  for \(y' = f(t, y)\),  ~\(y(t_0) = y_0\), \hb solution may or may not exist and 
solution may 
or may not be unique.}

But we have 2 theorems that guarantee both existence and uniqueness of solutions under certain conditions:
 

{\bf 1st order LINEAR differential equation:}
 

Thm 2.4.1:  If \(p:(a, b) \rightarrow R\) and \(g:(a, b) \rightarrow R\) are continuous and $a < 
t_0 < b$, then 
there exists a unique function \(y = \phi(t)\), \(\phi:(a, b) \rightarrow R\) that satisfies the 
initial value problem

\(y' + p(t) y = g(t)\), 

\(y(t_0) = y_0\)

{\bf 1st order differential equation (general case):}

Thm 2.4.2:  Suppose \(z = f(t, y)\) and \(z = {\partial f \over \partial y}(t, y)\) 
are continuous on $(a, b) 
point \((t_0, y_0) \in (a, b) \times (c, d)\), then there exists an interval $(t_0 - h, t_0 + h)  ubset 
(a, b)$ such that there exists a unique function 
\(y = \phi(t)\) 
defined on \((t_0 - h, t_0 + h)\) 
that 
satisfies the following
initial value problem:  \[y'  = f(t, y), ~~y(t_0) = y_0.\]

Note the initial value problem
   
 \[y' = y^{1\over 3}, ~y(0) = 0\] 
 
has an infinite number of different solutions.
\(y^{-{1\over 3}dy = dt\)}

\({3 \over 2y^{2 \over 3} = t + C\)}

\(y =\pm ({2 \over 3 t + C)^{3 \over 2}\)}

\(y(0) = 0\) implies \(C = 0\)

Thus \(y =\pm ({2 \over 3} t )^{3 \over 2}\) are solutions.

\(y = 0\) is also a solution, etc.

Compare to Thm 2.4.2: \hfil \break
 \(f(t, y) = y^{1 \over 3}\) is continuous near (0, 0)   \hfil \break
But \( {\partial f \over \partial y}(t, y) = {1 \over 3}y^{-2 \over 3}\) is not continuous near (0, 0) since 
it isn't defined at (0, 0).

{\bf Section 2.4 example:}  \({dy \over dt} = {1 \over (1 - t)(2 - y)}\)

\(F(y, t) =  {1 \over (1 - t)(2 - y)}\) is continuous for all \(t \not= 1\), \(y \not= 2\)

${\partial F \over \partial y} = {\partial  \left({1 \over (1 - t)(2 - y)}\right) \over \partial y} = 
 {1 \over (1 - t)}{\partial  (2 - y)^{-1} \over \partial y} 
 = {1 \over (1 - t)(2-y)^2}$

\({\partial F \over \partial y}\) is continuous for all \(t \not= 1\), \(y \not= 2\)

Thus the IVP \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(t_0) = y_0\) has a unique solution if \(t_0 \not= 1\), \(y_0 \not= 2\).

Note that if \(y_0 = 2\), \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(t_0) = 2\) has two solutions if \(t_0 \not= 1\) (and if we allow vertical slope in domain.  Note normally our convention will be to NOT allow vertical slope in domain of solution).

Note that if \(t_0 = 1\), \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(1) = y_0\) has no solutions.

\includegraphics[width=20ex]{domain}

 \((1,1/((1-t)(2-y)))/sqrt(1 + 1/((1-t)(2-y))^2)\) 

{\bf Solve via separation of variables:} \({dy \over dt} = {1 \over (1 - t)(2 - y)}\)

\(\int(2 - y)dy = \int {dt \over 1 - t}\)
implies
\(2y - {y^2 \over 2} = -ln|1 - t| + C\)

\(y^2 - 4y - 2ln|1 - t| + C = 0\)

\(y = {4 \pm  qrt{16 + 4(2ln|1 - t| + C)} \over 2}\)
\(= 2  \pm  qrt{4 + 2ln|1 - t| + C}\)
\(y= 2  \pm  qrt{2ln|1 - t| + C\)}

{\bf Find domain:} 
 \(2ln|1 - t| + C \geq 0\) \& \(t \not= 1\) \& \(y \not= 2\)

{\bf NOTE:  the convention in this class to to choose largest possible connected domain where tangent line to solution is never vertical.}

\(2ln|1 - t|  \geq -C\) and \(t \not= 1\) and \(y \not= 2\) implies

\(ln|1 - t| > -{C \over 2}\)  \hfill Note:  we want to find domain 
for this \(C\) 
{and thus this \(C\) can't swallow constants).}

 \(|1 - t| > e^{-{C \over 2}}\)  since \(e^x\) is an increasing function.

 \(1 - t < -e^{-{C \over 2}}\) or  \(1 - t > e^{-{C \over 2}}\)

 Domain: \(\cases{  t > e^{-{C \over 2}} + 1 & if \)t_0 > 1$ \cr
t < -e^{-{C \over 2}} + 1 &  if \(t_0 < 1\).}$

\end

START 34/2_4ex.tex

2.4 Solve Bernoulli's equation, 
\[y' + p(t)y = g(t)y^n,\] when \(n \not= 0, 1\) by changing it 
\[y^{-n}y' + p(t)y^{1-n} = g(t)\] when \(n \not = 0, 1\) by changing it 
to a linear equation by 
substituting \(v = y^{1-n}\)

Solve \(ty' + 2t^{-2}y = 2t^{-2}y^5\)

\(ty^{-5}y' + 2t^{-2}y^{-4} = 2t^{-2}\)

Let \(v = y^{-4}\).  Thus \(v' = -4y^{-5}y'\)

\(-4ty^{-5}y' - 8t^{-2}y^{-4} = -8t^{-2}\)

\(tv' - 8t^{-2}v = -8t^{-2}\)

Make coefficient of \(v' = 1\)

\(v' - 8t^{-3}v = -8t^{-3}\)

An antiderivative of \(-8t^{-3}\) is \(4t^{-2}\)

Multiply equation by \(e^{4t^{-2}}\)

\(e^{4t^{-2}}v' - 8t^{-3}e^{4t^{-2}}v = -8t^{-3}e^{4t^{-2}}\)

\((e^{4t^{-2}}v)' = -8t^{-3}e^{4t^{-2}}\) by PRODUCT rule.

\(\int (e^{4t^{-2}}v)' dt = -8\int t^{-3}e^{4t^{-2}} dt\)

\(e^{4t^{-2}}v = -8\int t^{-3}e^{4t^{-2}} dt\).  

Let \(u = 4t^{-2}\).  Then 
\(du = -8t^{-3}dt\)

\(e^{4t^{-2}}v = \int e^u du\)
\(= e^u + C\)

\(e^{4t^{-2}}v = e^{4t^{-2}} + C\)

\(v = 1 + C e^{-4t^{-2}}\)

\(y^{-4} = 1 + C e^{-4t^{-2}}\)
implies
\(y = \pm (1 + C e^{-4t^{-2}})^{-{1 \over 4}}\)

---

\(y' + {2 \over t-3}y = 1\)

An anti-derivative of \({2 \over t - 3} = 2 ln(t - 3)\)

\(e^{2ln(t - 3)} = e^{ln[(t-3)^2]} = (t - 3)^2\)

\(y' + {2 \over t-3}y = 1\)

\( (t - 3)^2y' + {2( t-3)}y =  (t - 3)^2\)

\(\int  [(t - 3)^2y]' =   \int (t - 3)^2\)

\((t - 3)^2 y =   {(t - 3)^3 \over 3} + C\)
implies
 \(y =  {(t - 3) \over 3} + C(t - 3)^{-2} \)

\end
---

\(y' + {2 \over t-3}y = e^t\)

An anti-derivative of ${2 \over t - 3} = 

\end

START 100/quiz4_2018ans.tex part 2

Quiz 4 ~~~SHOW ALL WORK\hfil \break
Nov 9, 2018

[15]~ 1.) Solve \(ty' + 4 y = t\) \hfill 

\(1y' + {4 \over t} y = 1\)

\(u(t) = e^{\int  {4\over t} dt } = e^{4ln| t|} = e^{ln(| t|^4)} = t^4\).

Let \(u(t) = t^4\)

\(t^4y' + {4 t^3} y = t^4\)

\((t^4 y) ' = t^4\) ~~~~~~~~~~{\bf Check this step:} \((t^4 y) ' = t^4y' + {4 t^3} y\)

\(\int (t^4 y) ' dt = \int t^4 dt\)

\(t^4 y = {t^5 \over 5} + C\)

\(y = {t \over 5} + Ct^{-4}\)

Answer:\(\underline{~~ y = {t \over 5} + Ct^{-4}~~}\)

\end

START 100/SPRING13/e1_2013ANS.txt part 2

2.)  Circle T for true and F for false.  Note that the answer to 2a is true.

[3]~ 2a.)   In more advanced math classes, you may be required to provide many more details when proving a function is onto.

\w

[4]~ 2b.)  Suppose 
\(\phi\)
is a solution to the equation, 
\(y' + p(t)y = g(t)\), 
then 
\(2\phi\) must also be a solution to
\(y' + p(t)y = g(t)\).

\w
[4]~ 2c.)  Suppose 
\(\phi\)
is a solution to the equation, 
\(y' + p(t)y^2 = 0\), 
then 
\(2\phi\) must also be a solution to
\(y' + p(t)y^2 = 0\).

\w

[4]~ 2d.)  Suppose 
\(\phi\)
is a solution to the equation, 
\(y' + p(t)y = 0\), 
then 
\(2\phi\) must also be a solution to
\(y' + p(t)y = 0\).

[15]~  4.) Solve the following initial value problem:  \(y'y = t + 3ty^2\), \(y(0) = -2\)

\(y'y = t + 3ty^2\)

\({dy \over dt} y = t(1 + 3y^2)\)

\({y dy \over 1 + 3y^2} = tdt\)

\(\int {y dy \over 1 + 3y^2} = \int tdt\)

\({1 \over 6} \int {6y dy \over 1 + 3y^2} = \int tdt\)

\({1 \over 6} ln|1 + 3y^2| = {1 \over 2} t^2 + C\)

\( ln|1 + 3y^2| = 3 t^2 + C\)

\( e^{ln|1 + 3y^2|} = e^{3 t^2 + C}\)

\(|1 + 3y^2| = e^{3 t^2}e^C\)

\(1 + 3y^2 = \pm e^C e^{3 t^2}\)

\(1 + 3y^2 = Ce^{3 t^2}\)

\( 3y^2 = Ce^{3 t^2} - 1\)

\( y^2 = {Ce^{3 t^2} - 1 \over 3}\)

\( y  = \pm  qrt

Callstack:
    at (Courses/University_of_Iowa/Differential_Equations_for_Engineers/04:_Problems_from_Math_2560_3600), /content/body/div[2]/div[4]/div[4]/p[22]/span, line 1, column 3

 \(y(0) = -2\): ~~   \( -2  = -  qrt

Callstack:
    at (Courses/University_of_Iowa/Differential_Equations_for_Engineers/04:_Problems_from_Math_2560_3600), /content/body/div[2]/div[4]/div[4]/p[23]/span[1], line 1, column 3

Callstack:
    at (Courses/University_of_Iowa/Differential_Equations_for_Engineers/04:_Problems_from_Math_2560_3600), /content/body/div[2]/div[4]/div[4]/p[23]/span[2], line 1, column 7

 \( 4  =

Callstack:
    at (Courses/University_of_Iowa/Differential_Equations_for_Engineers/04:_Problems_from_Math_2560_3600), /content/body/div[2]/div[4]/div[4]/p[24]/span, line 1, column 7

 fill
Answer 4.) \(\underline{~~ y  = -  qrt{{13e^{3 t^2 - 1 \over 3}}~~}\)}

\end

START 100/SPRING15/e1_2016ans.tex part 2

Math 3600 Differential Equations Exam \#1
                               March 2, 2016 \hfill  SHOW ALL
WORK
~~~

 2.)  Circle T for true and F for false.  

[4]~ 2a.)  The equation \(ln(t)y' = {t \over t+1} -  y(sin t^2)\) is a linear differential equation.

[4]~ 2b.)  The equation \(y' + y = y^2\) is a linear differential equation.
\hfill{~~~~~~~~~~~F}

[4]~ 2c.)  Suppose \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to 
 \(ay'' + by' + cy = 0\).
If \(y = h(t)\) is   also a solution to  \(ay'' + by' + cy = 0\), then there exists constants \(c_1\) and \(c_2\) such that \(h(t) = c_1\phi_1(t) + c_2\phi_2(t)\).
\hfill{~~~~~~~~~~~F}

[4]~ 2d.)   Suppose \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are linearly independent solutions to 
 \(ay'' + by' + cy = 0\).
If \(y = h(t)\) is   also a solution to  \(ay'' + by' + cy = 0\), then there exists constants \(c_1\) and \(c_2\) such that \(h(t) = c_1\phi_1(t) + c_2\phi_2(t)\).
\hfill{T~~~~~~~~~~~}

[4]~  3.)  By giving a specific counter-example, prove that \(y = ln(x)\) is
 not a linear function.

Proof 1:  \(ln(2) = ln(1 + 1) \not = 0  = ln(1) + ln(1)\).  

Proof 2:  \(ln(e) + ln(1) = 1 + 0 = 1\).  But \(ln(e+1) > ln(e) = 1\) since \(y = ln(x)\) is an increasing function (since \([ln(x)]' = {1 \over x} > 0\) for \(x > 0\)).

Proof 3:  \(ln(e) + ln(e) = 1 + 1 = 2\).  But \(ln(e +e) = ln(2e) = ln(2) + ln(e) = ln(2) + 1\).

\(ln(2) < ln(e) = 1\).  Thus \(2 \not= ln(2) + 1\).  Hence \(ln(e) + ln(e) \not= ln(2e)\) and thus \(y = ln(x)\) is not a linear function.

Proof 4:  \(e \cdot ln(1) = e \cdot 0 = 0 \not= 1 = ln(e) = ln(e \cdot 1) \).  In other words, if \(f(x) = ln(x) \), ~~\(f(cx) \not = cf(x)\) for constant \(c = e\) and \(x = 1\).

[20]~ 5.)  Solve:  
\(y' =  e^{4t} - {y \over t}\)

Linear, not separable:  \(y' + {y \over t} = e^{4t}\)

Let \(u(t) = e^{\int {dt \over t}} = e^{ln|t|+C} = C|t|\).  Let \(u(t) = t\)

 \(ty' + {y } = te^{4t}\)

\((ty)' =  te^{4t}\)

\(\int (ty)' = \int te^{4t} dt\).  

~~~~~~Let \(u = t\) ~~~ \(dv = e^{4t}\)

~~~~~~~~~~\(du = dt\)~~~ \(v = {e^{4t} \over 4}\)
  

\(ty = {te^{4t} \over 4} - \int {e^{4t} \over 4} dt = {te^{4t} \over 4} -  {e^{4t} \over 16} + C\)

Thus \(y = {e^{4t} \over 4} -  {e^{4t} \over 16t} + {C \over t}\)

 fill
Answer: \(\underline{~~y = {e^{4t \over 4} -  {e^{4t} \over 16t} + {C \over t}~~}\)}

\end

START 100/FALL18/final3600_F2016ans.tex part 1

MATH:3600:0002 Final Exam \hfil \break
Dec. 12, 2016 \hfill SHOW ALL STEPS \hfill

[14]~  2.)  Solve \(t \frac{dy}{dt} + y = 8 t^2\)

\(t y' + y = 8 t^2\) is a first order linear DE.

Short method:   

\([ty]' = t y' + y = 8 t^2\) 

\([ty]' = 8 t^2\) 

\(\int [ty]' = \int 8 t^2\)

\(ty = \frac{8}{3}t^3 + C\)

Thus \(y = \frac{8}{3}t^2 + Ct^{-1}\)

Longer method:

\( y' + \frac{1}{t}y = 8 t\)

Integrating factor:  \(u(t) = e^{\int \frac{dt}{t}} = e^{ln|t|} = |t|\)

Multiply both sides  by \(t\):  ~~ \(t y' + y = 8 t^2\) 
and continue as above.

 fill
Answer: \underbar{~~ \(y = \frac{8{3}t^2 + Ct^{-1}\)~~}}

\end

START 100/FALL16/e1_Fall2016ANS.tex part 3

[4]~  3.)  By giving a specific counter-example, prove that \(y = ln(x)\) is
 not a linear function.

[20]~ 5.)  Solve:  
\(y' =  e^{4t} - {y \over t}\)

 fill
Answer: \(\underline{~~\hskip 4in~~\)}

\end

START 100/ch1and2a.tex part 5

Section 2.5  Autonomous equations: \(y' = f(y)\)

If given either differential equation \(y' = f(y)\) 

 \rightline{ OR direction field:}
~~~~~~Find  equilibrium solutions and determine if 

stable, unstable, semi-stable.

~~~~~~Understand what the above means.

---
Asymptotically stable:
 fill fill
---
Asymptotically unstable:
 fill fill
---
Asymptotically semi-stable:

 fill

\end

\end

START 100/quiz2Fall2019Empty.tex part 3

[3]~ 2a.)  Sketch the direction field for the autonomous equation.  
{\(y' = ~~~~~~~~~~~~\)} 

 \includegraphics[width=30ex]{grapht-eps-converted-to}

[1]~ 2b.)  On the graph above, sketch the solution with initial value \(y(~~) = ~\)

[2]~ 2c.)  Find
the equilibrium solutions, and classify them as stable or unstable or semi-stable. 

 Equilibrium solution:  \(\underline{~~~~~~~~~~~~~~}\). ~ Stability of this equilibrium solution  \(\underline{~~~~~~~~~~~~~~~~~~~~}\).

 Equilibrium solution:  \(\underline{~~~~~~~~~~~~~~}\). ~ Stability of this equilibrium solution  \(\underline{~~~~~~~~~~~~~~~~~~~~}\).

---

From an old quiz:

1.  Sketch the direction field for the autonomous equation \(y' = y^2 + 2y - 8\). Find the equilibrium solutions, and classify them as stable
or unstable.  Sketch the solution with initial value \(y(0) = 1\). 
 \includegraphics[width=25ex]{grapht-eps-converted-to}

[2]~ Equilibrium solution:  \(\underline{~~~~~~~~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~~~~~~~~~~~~}\).

[2]~ Equilibrium solution:  \(\underline{~~~~~~~~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~~~~~~~~~~~~}\).

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 + 2y - 8\), \(y(0) = 3\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 + 2y - 8\), \(y(0) = 1\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

Answer to problem from old quiz:

1.  Sketch the direction field for the autonomous equation \(y' = y^2 + 2y - 8\). Find the equilibrium solutions, and classify them as stable
or unstable.  Sketch the solution with initial value \(y(0) = 1\). 

 \(y^2 + 2y - 8 = (y +4)(y-2) = 0\) implies \(y = -4\) and \(y = 2\) are equil solns.

[2]~ Equilibrium solution:  \(\underline{~~y = -4~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~stable~~}\).

[2]~ Equilibrium solution:  \(\underline{~~y = 2~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~unstable~~}\).

Note the equilibrium solution is a constant solution, not a number.  Thus I took off 0.5pt per problem if you did not include \(y =\) (i.e, 2 is incorrect, but the equation \(y = 2\) is correct).

---

[4]~  Sketch of Direction field and solution with initial value \(y(0) = 1\):

Note one really needs to draw slopes at 9 different \hb
\(y\)-values (Two for \(y > 4\), three for \(-2 < y < 4\), \hb and two for \(y < -2\) plus the 0-slopes at \(y = 4\) and\hb
 \(y = -2\)) in order to see all possible solutions.

Note, you don't need to calculate the slopes by \hb plugging 
in numbers.  You just need to know where \hb
slope is positive vs negative, small vs large. 

\end
2.)  Suppose \({\phi_n}\) is defined by sucessive approximation where  \(y' = {2t \over y+1}\), ~\(y(0) = 0\).
If \(\phi_1(t) = t^2\), then \(\phi_2(t) = \underline{~~ ln|t^2 + 1|~~}\)  \hfill (from 2.8)

\(y' = f(t, y(t))\), \(y(0) = 0\).  Thus \(y(t)  = \int_0^t f(s, y(s))ds\). 

\(\phi_2(t) = \int_0^t f(s, \phi_1(s))ds= \int_0^t {2s \over s^2 + 1} = ln|s^2 + 1| |_0^t  = ln|t^2 + 1| -  ln|1| =  ln|t^2 + 1| \)

\end{document}

START 100/quiz3.tex part 1

Match the following  general solution to the differential equation whose direction field is given below:

[5]~ \textbf{A)}  \(y = t + C\) \hfill ~~~~~
[5]~ \textbf{B)} \(y =2t + C\) \hfill ~\hfill ~ \hfill ~

[5]~ \textbf{C)}  \(y = \frac{1}{2}t + C\) \hfill ~~~~~
[5]~ \textbf{D)} \(y = -\frac{1}{2}t + C\) \hfill ~\hfill ~ \hfill ~

\leftline{\includegraphics[width = 2.2in, height = 2.2in]{SFminus1}}  %% export as pdf file 

\includegraphics[width = 2.2in, height = 2.2in]{SF2}  %% export as pdf file 

[30]~ 2.)  Answer both of the following questions.  If your proof to 2A is short, well-written, and correct, you will be given full credit for problem 2.  If your answer to 2A is incorrect or if it is not short and well-written (even if correct), your grade for problem 2 will depend solely on 2B.

2A.)  Prove that \(f: (0, \infty) \rightarrow R},~f(x) = ln(x)\)  is 1:1.

 fill

2B.)  Solve \(y' = y\).

 fill

p. 1 of 2 (\bf TURN PAGE)

Answer 2B: \underline{\hskip 3in

[50]~ 3.)  Solve the following differential equation (hint:  first get it into the appropriate format)  \]y' = \frac{y}{t} + ln(e)\[

 fill

Answer: \underline{\hskip 3in

\end{document}

4.)  Circle the general solution to the differential equation whose direction field is given below:

5.)  Which of the following could be the general solution to the differential equation whose direction field is given below:

6.)  Circle  the differential equation whose direction field is given below:

\end{document}

START 100/FALL17/quiz4Fall2017ans.tex part 1

1.  Sketch the direction field for the autonomous equation \(y' = y^2 + 2y - 8\). Find the equilibrium solutions, and classify them as stable
or unstable.  Sketch the solution with initial value \(y(0) = 1\). 

[2]~ Equilibrium solution:  \(\underline{\hskip 0.6in}\). ~~ Stability of this equilibrium solution  \(\underline{\hskip 0.6in}\).

[2]~ Equilibrium solution:  \(\underline{\hskip 0.6in}\). ~~ Stability of this equilibrium solution  \(\underline{\hskip 0.6in}\).

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' =  y^2 + 2y - 8\), \(y(0) = 1\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

[4]~  Sketch of Direction field and solution with initial value \(y(0) = 1\):

 fill

---

\end

START 100/FALL18/final3600_F2016ans.tex part 2

[14]~ 3.) Solve  \((3x^4 + 2y)dx + (2x +4y^3)dy= 0\)

\(M_y = 2\)  \(N_x = 2\)

If \(\psi_x = 3x^4 + 2y\), then \(\psi = \int(3x^4 + 2y)dx = \frac{3}{5}x^5 + 2xy + h(y)\).

\(\psi_y = 2x + h'(y) =2x +4y^3 \).  Thus \(h'(y) = 4y^3\) and \(h(y) = y^4\)

Answer:  \(\frac{3}{5}x^5 + 2xy + y^4 = C\)

Check:   \(\frac{d}{dx}(\frac{3}{5}x^5 + 2xy + y^4) = \frac{d}{dx}(C)\)

   \({3}x^4 + 2y + 2x\frac{dy}{dx} + 4y^3\frac{dy}{dx} = 0\)

 fill
Answer: \underbar{\hskip 5in}

\end

START 100/SPRING13/final3600ANS.tex part 2

[13]~ 2.)  Solve:  \((2x^3y^2 - 3x)dx + (x^4y - y^{-1}) dy = 0\)

Note:  $\frac{\partial}{\partial y} (2x^3y^2 - 3x) = 4x^3y
= \frac{\partial}{\partial x}(x^4y - y^{-1})$.  Thus equation is exact.

Let \(\psi_x = 2x^3y^2 - 3x\).  Then \(\psi = \frac{x^4y^2 - 3x^2}{2} + h(y)\)
~~and~~
\(\psi_y = x^4y + h'(y) = x^4y - y^{-1}\)

\(\int h'(y)dy = \int -y^{-1}dy = -ln|y|\) 
~~and~~
\( \frac{x^4y^2 - 3x^2}{2} -ln|y| = C\)

\end

START 100/FALL17/quiz3Fall2017formAans.tex part 1

\nopagenumbers

Quiz 3 Form A \hfil \break
Sept 25, 2017

1.  Sketch the direction field for the autonomous equation \(y' = y^2 + 2y - 8\). Find the equilibrium solutions, and classify them as stable
or unstable.  Sketch the solution with initial value \(y(0) = 1\). 

 \(y^2 + 2y - 8 = (y +4)(y-2) = 0\) implies \(y = -4\) and \(y = 2\) are equil solns.

[2]~ Equilibrium solution:  \(\underline{~~y = -4~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~stable~~}\).

[2]~ Equilibrium solution:  \(\underline{~~y = 2~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~unstable~~}\).

Note the equilibrium solution is a constant solution, not a number.  Thus I took off 0.5pt per problem if you did not include \(y =\) (i.e, 2 is incorrect, but the equation \(y = 2\) is correct).

---

---

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 - 5y - 6\), \(y(0) = 3\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 \(y^2 - 5y - 6 = (y - 6)(y+1) = 0\) implies \(y = 6\) and \(y = -1\) are equil solns. % Note \(y = 3 \in (-1, 6)\) %and \(y = -1\) is a stable equil
\(y \rightarrow -1\) as \(t \rightarrow +\infty\) 

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 + 2y - 8\), \(y(0) = 3\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 Note \(y(0) = 3 > 2\).  Thus 
\(y \rightarrow +\infty\) as \(t \rightarrow +\infty\) 

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 + 2y - 8\), \(y(0) = 1\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 Note \(y(0) = 1 \in (-4, 2)\).  Thus 
\(y \rightarrow -2\) as \(t \rightarrow +\infty\) 

---

[4]~  Sketch of Direction field and solution with initial value \(y(0) = 3\):

Note one really needs to draw slopes at 9 different \hb
\(y\)-values (Two for \(y > 4\), three for \(-2 < y < 4\), \hb and two for \(y < -2\) plus the 0-slopes at \(y = 4\) and\hb
 \(y = -2\)) in order to see all possible solutions.

Note, you don't need to calculate the slopes by \hb plugging 
in numbers.  You just need to know where \hb
slope is positive vs negative, small vs large. 

Choose the best answer for the following problems:

[2]~ 2.)  If \(u = y - x\), then  E.)  \({dy \over dx} = {du \over dx} +1\)  

\(y = u + x\) \& hence \({dy \over dx} = {d \over dx}(u+x) = {du \over dx} + {dx \over dx} = {du \over dx} +1\), since derivative is a linear fn.

---

[2]~ 3.)  The integrating factor used to solve the differential equation \({dy \over dx} - {y \over x} = x^2\) is

{D.)  \({1 \over x}\) }\hfill
since \(u(x) = e^{\int{ -1 \over x}dx} = e^{-ln|x|} = e^{ln|x|^{-1}} = |x|^{-1}\)

Check:  If \(u(x) = {1 \over x} = x^{-1}\), then 

---

[2]~ 4.)  The solution to the initial value problem \(y' = {1 - x \over y}\), \(y(0) = - qrt{3}\) is\hb  \(y = -  qrt{-x^2 + 2x + 3}\).~~
State the largest interval on which the solution is defined.

\(-x^2 + 2x + 3 \geq 0\) and since our convention is to not allow points vertical tangent lines in our domain, \(y \not= 0\).

Thus \(-x^2 + 2x + 3 = (-x + 3)(x+1) > 0\).  Hence either \((-x + 3)\) and \((x+1)\)
are both positive or both negative.  Thus

\(-x + 3 > 0\) and \(x+1 > 0\) and thus  \(3 > x\) and \(x > -1\).  I.e., \(x \in (-1, 3)\) 
~~~~~OR
\(-x + 3 < 0\) and \(x+1 < 0\) and thus  \(3 < x\) and \(x < -1\).  But \(x\) can't be both greater than 3 and less than -1, so largest possible  domain is ~~~B.)~ (-1, 3).

---

[2]~ 5.)  A tank contains 100 liters of pure water. Saline solution with a variable
concentration \(c(t) = e^{-t \over 100}\) grams of salt per liter is pumped into the
tank at rate 1 liter per minute. The tank is kept perfectly mixed and
also drains at a rate of 1 liter per minute, so the volume stays constant.
Write an initial value problem modeling \(A(t)\), the amount of grams of
salt dissolved in the solution in the tank at time \(t\) minutes.

Note since the water is pure, the tank starts off with 0 grams of water.  Thus \(A(0) = 0\).  
{Rate in = \({e^{-t \over 100} g \over 1 L}{1 L \over 1 minute}\)
\hfill
Rate out = \({A(t) g \over 1 L}{1 L \over 1 minute}\)
\hfill

Thus~~
A.)  \({dA \over dt} =  e^{-t \over 100} - {A \over 100}\), ~
\(A(0) = 0\)}

\end

START 100/FALL17/quiz3Fall2017formBans.tex part 1

\nopagenumbers

Quiz 3 Form B \hfil \break
Sept 25, 2017

1.  Sketch the direction field for the autonomous equation \(y' = y^2 - 2y - 8\). Find the equilibrium solutions, and classify them as stable
or unstable.  Sketch the solution with initial value \(y(0) = 1\). 

 \(y^2 - 2y - 8 = (y +4)(y-2) = 0\) implies \(y = 4\) and \(y = -2\) are equil solns.

[2]~ Equilibrium solution:  \(\underline{~~y = 4~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~unstable~~}\).

[2]~ Equilibrium solution:  \(\underline{~~y = -2~~}\). ~~ Stability of this equilibrium solution  \(\underline{~~stable~~}\).

Note the equilibrium solution is a constant solution, not a number.  Thus I took off 0.5pt per problem if you did not include \(y =\) (i.e, 4 is incorrect, but the equation \(y = 4\) is correct).

---

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 - 5y - 6\), \(y(0) = 3\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 \(y^2 - 5y - 6 = (y - 6)(y+1) = 0\) implies \(y = 6\) and \(y = -1\) are equil solns. % Note \(y = 3 \in (-1, 6)\) %and \(y = -1\) is a stable equil
\(y \rightarrow -1\) as \(t \rightarrow +\infty\) 

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 - 2y - 8\), \(y(0) = 5\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 Note \(y(0) = 5 > 4\).  Thus 
\(y \rightarrow +\infty\) as \(t \rightarrow +\infty\) 

[2]~ If \(y  = \phi(t)\) is the solution to the initial value problem \(y' = y^2 - 2y - 8\), \(y(0) = 1\), what happens to \(y = \phi(t)\) as \(t\) goes to \(+\infty\)?

 Note \(y(0) = 1 \in (-4, 2)\).  Thus 
\(y \rightarrow -2\) as \(t \rightarrow +\infty\) 

---

[4]~  Sketch of Direction field and solution with initial value \(y(0) = 3\):

Note one really needs to draw slopes at 9 different \hb
\(y\)-values (Two for \(y > 4\), three for \(-2 < y < 4\), \hb and two for \(y < -2\) plus the 0-slopes at \(y = 4\) and\hb
 \(y = -2\)) in order to see all possible solutions.

Note, you don't need to calculate the slopes by \hb plugging 
in numbers.  You just need to know where \hb
slope is positive vs negative, small vs large. 

Choose the best answer for the following problems:

[2]~ 2.)  If \(u = y - x\), then  E.)  \({dy \over dx} = {du \over dx} +1\)  

\(y = u + x\) \& hence \({dy \over dx} = {d \over dx}(u+x) = {du \over dx} + {dx \over dx} = {du \over dx} +1\), since derivative is a linear fn.

---

[2]~ 3.)  The integrating factor used to solve the differential equation \({dy \over dx} - {y \over x} = x^2\) is

{D.)  \({1 \over x}\) }\hfill
since \(u(x) = e^{\int{ -1 \over x}dx} = e^{-ln|x|} = e^{ln|x|^{-1}} = |x|^{-1}\)

Check:  If \(u(x) = {1 \over x} = x^{-1}\), then 

---

[2]~ 4.)  The solution to the initial value problem \(y' = {1 - x \over y}\), \(y(0) = - qrt{3}\) is\hb  \(y = -  qrt{-x^2 + 2x + 3}\).~~
State the largest interval on which the solution is defined.

\(-x^2 + 2x + 3 \geq 0\) and since our convention is to not allow points vertical tangent lines in our domain, \(y \not= 0\).

Thus \(-x^2 + 2x + 3 = (-x + 3)(x+1) > 0\).  Hence either \((-x + 3)\) and \((x+1)\)
are both positive or both negative.  Thus

\(-x + 3 > 0\) and \(x+1 > 0\) and thus  \(3 > x\) and \(x > -1\).  I.e., \(x \in (-1, 3)\) 
~~~~~OR
\(-x + 3 < 0\) and \(x+1 < 0\) and thus  \(3 < x\) and \(x < -1\).  But \(x\) can't be both greater than 3 and less than -1, so largest possible  domain is ~~~B.)~ (-1, 3).

---

[2]~ 5.)  A tank contains 100 liters of pure water. Saline solution with a variable
concentration \(c(t) = e^{-t \over 100}\) grams of salt per liter is pumped into the
tank at rate 1 liter per minute. The tank is kept perfectly mixed and
also drains at a rate of 1 liter per minute, so the volume stays constant.
Write an initial value problem modeling \(A(t)\), the amount of grams of
salt dissolved in the solution in the tank at time \(t\) minutes.

Note since the water is pure, the tank starts off with 0 grams of water.  Thus \(A(0) = 0\).  
{Rate in = \({e^{-t \over 100} g \over 1 L}{1 L \over 1 minute}\)
\hfill
Rate out = \({A(t) g \over 1 L}{1 L \over 1 minute}\)
\hfill

Thus~~
A.)  \({dA \over dt} =  e^{-t \over 100} - {A \over 100}\), ~
\(A(0) = 0\)}

---

\end

START 100/FALL17/quiz4Fall2017ans.tex part 2

Choose the best answer for the following problems:

[2]~ 2.)  If \(u = y - x\), then

A.)  \({dy \over dx} = {du \over dx}\)  \hfill
B.)  \({dy \over dx} = {dx \over du}\)  \hfill
C.)  \({dy \over dx} = {dx \over dy}\)  \hfill

D.)  \({dy \over dx} = {du \over dy}\)  \hfill
E.)  \({dy \over dx} = {du \over dx} + 1\)  \hfill
F.)  \({dy \over dx} = {du \over dx}+ {dx \over du}\)  \hfill

[2]~ 3.)  The integrating factor used to solve the differential equation \({dy \over dx} - {y \over x} = x^2\) is

A.)  \(e^{x}\) \hfill
B.)  \(e^{-x}\) \hfill
C.)  \({x}\) \hfill
D.)  \({1 \over x}\) \hfill

---

[2]~ 4.)  The solution to the initial value problem \(y' = {1 - x \over y}\), \(y(0) = - qrt{3}\) is\hb  \(y = -  qrt{-x^2 + 2x + 3}\).~~
State the largest interval on which the solution is defined.

A.) ~ \((-1, \infty)\) \hfill
B.)  ~\((-1, 3)\) \hfill
C.)  ~\((3, \infty)\) \hfill
D.)  ~\((-\infty, 1)\) \hfill
E.)  ~\((-\infty, 3)\) \hfill

F.) ~ \([-1, \infty)\) \hfill
G.)  ~\([-1, 3]\) \hfill
H.)  ~\([3, \infty)\) \hfill
I.)  ~\((-\infty, 1]\) \hfill
J.)  ~\((-\infty, 3]\) \hfill

\end

START 100/2_5a.tex

{\bf Section 2.4 example:}  \({dy \over dt} = {1 \over (1 - t)(2 - y)}\)

\(F(y, t) =  {1 \over (1 - t)(2 - y)}\) is continuous for all \(t \not= 1\), \(y \not= 2\)

${\partial F \over \partial y} = {\partial  \left({1 \over (1 - t)(2 - y)}\right) \over \partial y} = 
 {1 \over (1 - t)}{\partial  (2 - y)^{-1} \over \partial y} 
 = {1 \over (1 - t)(2-y)^2}$

\({\partial F \over \partial y}\) is continuous for all \(t \not= 1\), \(y \not= 2\)

Thus the IVP \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(t_0) = y_0\) has a unique solution if \(t_0 \not= 1\), \(y_0 \not= 2\).

Note that if \(y_0 = 2\), \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(t_0) = 2\) has two solutions if \(t_0 \not= 2\)

Note that if \(t_0 = 1\), \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(1) = y_0\) has no solutions.

 \((1,1/((1-x)(2-y)))/sqrt{1 + 1/((1-x)(2-y))^2)\) }

{\bf Solve via separation of variables:}

\(\int(2 - y)dy = \int {dt \over 1 - t}\)

\(2y - {y^2 \over 2} = -ln|1 - t| + C\)

\(y^2 - 4y - 2ln|1 - t| + C = 0\)

\(y = {4 \pm  qrt{16 + 4(2ln|1 - t| + C)} \over 2}\)
\(= 2  \pm  qrt{4 + 2ln|1 - t| + C}\)
\(y= 2  \pm  qrt{2ln|1 - t| + C\)}

{\bf Find domain:}  \(2ln|1 - t| + C \geq 0\)

\(2ln|1 - t|  \geq -C\)

\(ln|1 - t| \geq -{C \over 2}\)  \hfill Note:  we want to find domain 
for this \(C\) 

 \(|1 - t| \geq e^{-{C \over 2}}\)  since \(e^x\) is an increasing function.

 \(1 - t \leq -e^{-{C \over 2}}\) or  \(1 - t \geq e^{-{C \over 2}}\)

 \(- t \leq -e^{-{C \over 2}} - 1\) or  \(-t \geq e^{-{C \over 2}}- 1\)

 Domain: \(\cases{  t \geq e^{-{C \over 2}} + 1 & if \)t_0 > 0$ \cr
t \leq -e^{-{C \over 2}} + 1 &  if \(t_0 < 0\).}$

Note:  Domain is much easier to determine when the ODE is linear.

 
{\bf Find C given \(y(t_0) = y_0\):}  \(y_0 =  2  \pm  qrt{2ln|1 - t_0| + C}\)

\(\pm(y_0 -2) =     qrt{2ln|1 - t_0| + C}\)

\((y_0 -2)^2 - 2ln|1 - t_0| = C\)

\(y = 2  \pm  qrt{2ln|1 - t| + C}\)

\(y = 2  \pm  qrt{2ln|1 - t| + (y_0 -2)^2 - 2ln|1 - t_0|}\)
 
\(y = 2  \pm  qrt{ (y_0 -2)^2 + ln{(1 - t)^2 \over (1-t_0)^2}}\)

 Domain: \(\cases{  t \geq e^{-{C \over 2}} + 1 & if \)t_0 > 0$ \cr
t \leq -e^{-{C \over 2}} + 1 &  if \(t_0 < 0\).}$

$e^{-{C \over 2}} = e^{-{(y_0 -2)^2 - 2ln|1 - t_0| \over 2}}
= |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} }$

 Domain: \(\cases{  t \geq 1 + |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} } & if \)t_0 > 0$ \cr
t \leq 1 - |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} } &  if \(t_0 < 0\).}$

---
Section 2.5:  

Exponential Growth/Decay \hfill \break
Example:  population growth/radioactive decay)

\(y' = ry\), \(y(0) = y_0\) implies \(y = y_0 e^{rt}\)

\(r > 0\) \hskip 1.2in \(r < 0\)

Logistic growth:  \(y' = h(y)y\)

Example:  \(y' = r(1 - {y \over K})y\)

\(y\) vs \(f(y)\) \hskip 1in slope field:

Equilibrium solutions:

Asymptotically stable:
 fill
Asymptotically unstable:
 fill
Asymptotically semi-stable:

 fill
As \(t \rightarrow \infty\), if \(y > 0\), \(y \rightarrow \)
 fill

Solution:  \(y = {y_0K \over y_0 + (K - y_0)e^{-rt}}\)

{\bf Section 2.7 Euler method:}  Using tangent lines to approximate a function. 

 
\(y_{i+1} = y_i + \Delta y = y_i + {\Delta y \over \Delta t}\Delta t \cong  y_i + {dy \over dt}\Delta t\)

Alternatively use equation of tangent line: 

 slope \(= {y_{i+1 - y_i \over t_{i+1} - t_i} = f'(y_i, t_i)\).}

 
\(y_{i+1} = f'(y_i, t_i)( t_{i+1} - t_i) + y_i = T(t_{i+1})\) where \(y = T(t)\) 

 
Example:  \({dy \over dt} = y^2\), \(y(2) = 1\) implies 
\(y = {1 \over 3 - t}\). 

 box{ \offinterlineskip \halign{  trut 
 rule \hfil  ~# \hfil 
& rule width 1.25 pt \hfil \hskip 0.1in~# \hskip 0.1in\hfil
& rule width 1.25 pt\hfil \hskip 0.2in~#\hskip 0.2in \hfil
 rule \cr
\noalign{---ule}
\(t\) & \(y = 1/(3-t)\)& approximation \cr 
\noalign{---ule height 1.5pt}
2.000000 & 1.000000 & 1.000000 \cr 
 \noalign{ ---ule}
3.000000 & 999.000000 & 2.000000 \cr 
 \noalign{ ---ule}
4.000000 & -1.000000 & 6.000000 \cr 
 \noalign{ ---ule}
5.000000 & -0.500000 & 42.000000 \cr 
 \noalign{ ---ule}
6.000000 & -0.333333 & 1806.000000 \cr 
 \noalign{ ---ule}
 %\noalign{ ---ule}
 %\noalign{ ---ule}
 %\noalign{ ---ule}
 %\noalign{ ---ule}
}
}

 fill

 box{ \offinterlineskip \halign{  trut 
 rule \hfil  ~# \hfil 
& rule width 1.25 pt \hfil \hskip 0.1in~# \hskip 0.1in\hfil
& rule width 1.25 pt\hfil \hskip 0.2in~#\hskip 0.2in \hfil
 rule \cr
\noalign{---ule}
\(t\) & \(y = 1/(3-t)\)& approximation \cr 
\noalign{---ule height 1.5pt}
2.000000 & 1.000000 & 1.000000 \cr 
 \noalign{ ---ule}
2.100000 & 1.111111 & 1.100000 \cr 
 \noalign{ ---ule}
2.200000 & 1.250000 & 1.221000 \cr 
 \noalign{ ---ule}
2.300000 & 1.428571 & 1.370084 \cr 
 \noalign{ ---ule}
2.400000 & 1.666667 & 1.557797 \cr 
 \noalign{ ---ule}
2.500000 & 2.000000 & 1.800470 \cr 
 \noalign{ ---ule}
2.600000 & 2.500000 & 2.124640 \cr 
 \noalign{ ---ule}
2.700000 & 3.333333 & 2.576049 \cr 
 \noalign{ ---ule}
2.800000 & 5.000000 & 3.239652 \cr 
 \noalign{ ---ule}
2.900000 & 10.000004 & 4.289186 \cr 
 \noalign{ ---ule}
}
}

 box{ \offinterlineskip \halign{  trut 
 rule \hfil  ~# \hfil 
& rule width 1.25 pt  \hfil  ~# \hfil 
& rule width 1.25 pt\ \hfil  ~# \hfil 
 rule \cr
\noalign{---ule}
\(t\) & \(y = 1/(3-t)\)& approximation \cr 
\noalign{---ule height 1.5pt}
2.00 & 1.000000 & 1.000000 \cr 
 \noalign{ ---ule}
2.01 & 1.010101 & 1.010000 \cr 
 \noalign{ ---ule}
2.02 & 1.020408 & 1.020201 \cr 
 \noalign{ ---ule}
2.03 & 1.030928 & 1.030609 \cr 
 \noalign{ ---ule}
2.04 & 1.041667 & 1.041231 \cr 
 \noalign{ ---ule}
2.05 & 1.052632 & 1.052072 \cr 
 \noalign{ ---ule}
2.06 & 1.063830 & 1.063141 \cr 
 \noalign{ ---ule}
2.07 & 1.075269 & 1.074443 \cr 
 \noalign{ ---ule}
2.08 & 1.086957 & 1.085988 \cr 
 \noalign{ ---ule}
2.09 & 1.098901 & 1.097782 \cr 
 \noalign{ ---ule}
2.10 & 1.111111 & 1.109833 \cr 
 \noalign{ ---ule}
2.11 & 1.123595 & 1.122150 \cr 
 \noalign{ ---ule}
2.12 & 1.136364 & 1.134742 \cr 
 \noalign{ ---ule}
2.13 & 1.149425 & 1.147619 \cr 
 \noalign{ ---ule}
 \noalign{ ---ule}
 \noalign{ ---ule}
 \noalign{ ---ule}
2.87 & 7.692308 & 6.721314 \cr 
 \noalign{ ---ule}
2.88 & 8.333333 & 7.173075 \cr 
 \noalign{ ---ule}
2.89 & 9.090908 & 7.687605 \cr 
 \noalign{ ---ule}
2.90 & 9.999998 & 8.278598 \cr 
 \noalign{ ---ule}
2.91 & 11.111107 & 8.963949 \cr 
 \noalign{ ---ule}
2.92 & 12.499993 & 9.767473 \cr 
 \noalign{ ---ule}
2.93 & 14.285716 & 10.721509 \cr 
 \noalign{ ---ule}
2.94 & 16.666666 & 11.871017 \cr 
 \noalign{ ---ule}
2.95 & 19.999996 & 13.280227 \cr 
 \noalign{ ---ule}
2.96 & 24.999987 & 15.043871 \cr 
 \noalign{ ---ule}
2.97 & 33.333298 & 17.307051 \cr 
 \noalign{ ---ule}
2.98 & 49.999897 & 20.302391 \cr 
 \noalign{ ---ule}
2.99 & 99.999496 & 24.424261 \cr 
 \noalign{ ---ule}
}
}

\end

$\matrix{ t & y & approximation \cr 
2 & 1 & 1}$

\((1,1/((1-x)(2+x)))/sqrt{1 + 1/((1-x)(2+x))^2\)

\({1,(1-x)(2+x)}/sqrt{1 + ((1-x)(2+x))^2}\)

\({1,(1-y)(2+y)}/sqrt{1 + ((1-y)(2+y))^2}\)

\end

float t;
float y;
float yAprox;
float h = 0.1;

for (i = 0; i < 10; i++ )
{
t = 2 + i*h;
y = 1/(3 - 2 + i*h);
yAprox = yAprox +  yAprox*yAprox*h
printf( ``%d & %d & %d'', t, y, yAprox );
}

START 100/2_8new.tex

2.8:  Approximating solution using 

\bf Method of Successive Approximation
\u
(also called Picard's iteration method).

IVP:   \(y' = f(t, y)\), \(y(t_0) = y_0\).

Note: Can always translate IVP to move initial value to the origin and translate back after solving:

Hence for simplicity {\bf in section 2.8}, we will assume  initial value is at the origin:    \(y' = f(t, y)\), \(y(0) = 0\).

{\bf Thm 2.4.2:}  Suppose the functions \hfill \break 
\(z = f(t, y)\) and \(z = {\partial f \over \partial y}(t, y)\) 
are continuous on \hb
\((a, b) \times (c, d)\) 
and the      
point \((t_0, y_0) \in (a, b) \times (c, d)\), 
then there exists an interval $(t_0 - h, t_0 + h)  ubset 
(a, b)$ such that there exists a unique function 
\(y = \phi(t)\) 
defined on \((t_0 - h, t_0 + h)\) 
that 
satisfies the following
initial value problem:  
\(y'  = f(t, y), ~~y(t_0) = y_0.\)

Thm 2.8.1 is translated to  origin version of Thm 2.4.2:

{\bf Thm 2.8.1:}  Suppose the functions \hfill \break 
\(z = f(t, y)\) and \(z = {\partial f \over \partial y}(t, y)\) 
are continuous for all \(t\) in 
\((-a, a) \times (-c, c)\), 

\u
then there exists an interval $(- h,  h)  ubset 
(-a, a)$ such that there exists a unique function 
\(y = \phi(t)\) 
defined on \((- h, h)\) 
that 
satisfies the following
initial value problem:  
 
\(y'  = f(t, y), ~~y(0) = 0.\)
{\bf Proof outline} (note this is a constructive proof and thus the proof is very useful).

Given:  \(y' = f(t, y)\), \(y(0) = 0\)  \hfil Eqn (*)
\u
\(f\), \(\partial f/\partial y\) continuous \(\forall (t, y) \in (-a, a) \times (-b, b)\).  

Then
\(y = \phi(t)\) is a solution to (*) iff

\(\p'(t) = f(t, \p(t))\), ~~\(\p(0) = 0\) iff

 \(\int_0^t \p'(s) ds = \i f(s, \phi(s)) ds\), ~~\(\p(0) = 0\) iff
 
\( \p(t) = \p(t) - \p(0) =   \i f(s, \phi(s)) ds\) 

Thus \(y = \p(t)\) is a solution to (*) 

Construct \(\phi\) using method of successive approximation --  also called Picard's iteration method.

Let \(\phi_0(t) = 0\) (or the function of your choice)

Let \(\p_1(t) =   \i f(s, \phi_0(s)) ds\) 

Let \(\p_2(t) =   \i f(s, \phi_1(s)) ds\) 

\( dots\)

Let \(\p_{n+1}(t) =   \i f(s, \phi_n(s)) ds\) 

Let \(\phi(t) = \lim_{n \rightarrow \infty} \p_n(t)\)

To finish the proof, need to answer the following questions (see book or more advanced class):

1.)  Does \(\p_n(t)\) exist for all \(n\)?

2.)  Does sequence \(\phi_n\) converge?

3.)  Is \(\phi(t) = \lim_{n \rightarrow \infty} \p_n(t)\) a solution to (*).

4.)  Is the solution unique.

Example:  \(y' = t + 2y\).  \hfil That is \(f(t, y) = t + 2y\)

Let \(\phi_0(t) = 0\)

Let \(\p_1(t) =   \i f(s, 0)ds = \i (s + 2(0))ds\)

Let \(\p_2(t) =   \i f(s, \p_1(s))ds =   \i f(s, {s^2 \over 2})ds\)

Let \(\p_3(t) =   \i f(s, \p_2(s))ds =   \i f(s, {s^2 \over 2}+  {s^3 \over 3})ds\)

Let \(\p_4(t) =   \i f(s, \p_3(s))ds   \)

\hskip 0.59in \(= \i f(s, {s^2 \over 2}+  {s^3 \over 3}+ {s^4 \over 6} )ds\)

\hskip 0.59in \( = \i (s + 2({s^2 \over 2}+  {s^3 \over 3} + {s^4 \over 6}))ds\)

\hskip 0.59in\( =  {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}\)

\( dots\)

Determine formula for \(\p_n\):

Note patterns:

\(\i s ds= {t^2 \over 2} = \)

\(\i {s^2 \over 2}ds = {t^3 \over 3 \cdot 2} = \)

$\i {s^3 \over 3 \cdot 2}ds = {t^4 \over 4 \cdot 
3 \cdot 2} = $

$\i {s^4 \over 4 \cdot 3 \cdot 2}ds = {t^5 \over 5 \cdot 4 \cdot 
3 \cdot 2} = $

Thus look for factorials.

 \(\phi_0(t) = 0\)

 \(\p_1(t) = {t^2 \over 2} \) 

 \(\p_2(t)  =   {t^2 \over 2} +  {t^3 \over 3}\)

 \(\p_3(t) =    {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6}\)

 \(\p_4(t) =   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}=   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 3 \cdot 2} + {t^5 \over 5 \cdot 3}\)

 fill

Thus \(\p_n(t) =\)

FYI (ie not on quizzes/exam):
Defn:  \(\So a_kx ^k = \displaystyle\lim_{n \rightarrow \infty}  o a_kx^k\)

{Taylor's Theorem}:  If \(f\) is analytic  at 0, then for small \(x\) (i.e., \(x\) near 0), 
\(f(x)=\displaystyle um_{k=0^\infty{f^{(k)}(0) \over k!}x^k\) \hskip 1.13in}
 
Example:
 
\(e^t = \So {t^k \over k!}\)  and thus \(e^{bt} = \So {b^kt^k \over k!}\) for \(t\) near 0.

\(\p_n(t) = \displaystyle um_{k = 2}^n {2^{k-2} \over k!} t^k\)
Thus $\p(t) = \displaystyle\lim_{n \rightarrow \infty}\p_n(t) = 
\displaystyle um_{k = 2}^\infty {2^{k-2} \over k!} t^k
 = 
{1 \over 4}\displaystyle um_{k = 2}^\infty {2^{k} \over k!} t^k$
 \rightline{= \( \displaystyle{1 \over 4}\displaystyle\left(~~~~~~~~~~~- ~~~~~~~~~ - ~~~~~~~~~\displaystyle\right)\)}

2.8:  Approximating soln to IVP using seq of fns.

\(\phi_0(t) = 0\), ~~ \textcolor{red}{\(\p_1(t) = {t^2 \over 2} \)},~~
  \textcolor{magenta}{\(\p_2(t)  =   {t^2 \over 2} +  {t^3 \over 3}\)},

 \(\p_4(t) =   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}\)

2.7: Approximating soln to IVP using multiple tangent lines.

$y(t) = \cases{ 
0 & \(0 \leq t \leq 0.1\) \cr
0.1t -0.01  & \(0.1 \leq t \leq 0.2\) \cr
0.22t -0.034 & \(0.2 \leq t \leq 0.3\) \cr
0.364t -0.0772 & \(0.3 \leq t \leq 0.4\) \cr
0.5328t -0.14672 & \(0.4 \leq t \leq 0.5\) \cr
$}

\end

START 100/exam1review.tex part 4

2.7: Approximating soln to IVP using multiple tangent lines.

Example:    \(y' = t + 2y\), \(y(0) = 0\)

$y(t) = \cases{ 
0 & \(0 \leq t \leq 0.1\) \cr
0.1t -0.01  & \(0.1 \leq t \leq 0.2\) \cr
0.22t -0.034 & \(0.2 \leq t \leq 0.3\) \cr
0.364t -0.0772 & \(0.3 \leq t \leq 0.4\) \cr
0.5328t -0.14672 & \(0.4 \leq t \leq 0.5\) \cr
$}

~~~\includegraphics[width = 6.4truein]{fig2_7.png}

2.8:  Approximating soln to IVP using seq of fns, 
\(\p_{n+1(t) =   \i f(s, \phi_n(s)) ds\) }

Example:    \(y' = t + 2y\), \(y(0) = 0\)

\(\phi_0(t) = 0\), ~~ \textcolor{red}{\(\p_1(t) = {t^2 \over 2} \)},~~
  \textcolor{magenta}{\(\p_2(t)  =   {t^2 \over 2} +  {t^3 \over 3}\)},

 \(\p_4(t) =   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}\)

~~~\includegraphics[width = 6.3truein]{Figure2_8.png}

\end

START 100/quiz2Fall2019Empty.tex part 4

---

1.) When taking the derivative with respect to \(t\),  \((y^2)' = 2y\)

\hfill   A)  True  \hfill  B)False \hfill ~~

2.) When taking the derivative with respect to \(t\),  \((y^2)' = 2yy'\)    

\hfill   A)  True  \hfill  B)False \hfill ~~

3.) If \(y(0) = -2\) and \(y^2 = g(t)\),  then \(y(t) =  qrt{g(t)}\)

\hfill   A)  True  \hfill  B)False \hfill ~~

4.) If \(y(0) = -2\) and \(y^2 = g(t)\),  then \(y(t) = - qrt{g(t)}\)

\hfill   A)  True  \hfill  B)False \hfill ~~

\end

START 100/2_5a.tex

{\bf Section 2.4 example:}  \({dy \over dt} = {1 \over (1 - t)(2 - y)}\)

\(F(y, t) =  {1 \over (1 - t)(2 - y)}\) is continuous for all \(t \not= 1\), \(y \not= 2\)

${\partial F \over \partial y} = {\partial  \left({1 \over (1 - t)(2 - y)}\right) \over \partial y} = 
 {1 \over (1 - t)}{\partial  (2 - y)^{-1} \over \partial y} 
 = {1 \over (1 - t)(2-y)^2}$

\({\partial F \over \partial y}\) is continuous for all \(t \not= 1\), \(y \not= 2\)

Thus the IVP \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(t_0) = y_0\) has a unique solution if \(t_0 \not= 1\), \(y_0 \not= 2\).

Note that if \(y_0 = 2\), \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(t_0) = 2\) has two solutions if \(t_0 \not= 2\)

Note that if \(t_0 = 1\), \({dy \over dt} = {1 \over (1 - t)(2 - y)}\), \(y(1) = y_0\) has no solutions.

 \((1,1/((1-x)(2-y)))/sqrt{1 + 1/((1-x)(2-y))^2)\) }

{\bf Solve via separation of variables:}

\(\int(2 - y)dy = \int {dt \over 1 - t}\)

\(2y - {y^2 \over 2} = -ln|1 - t| + C\)

\(y^2 - 4y - 2ln|1 - t| + C = 0\)

\(y = {4 \pm  qrt{16 + 4(2ln|1 - t| + C)} \over 2}\)
\(= 2  \pm  qrt{4 + 2ln|1 - t| + C}\)
\(y= 2  \pm  qrt{2ln|1 - t| + C\)}

{\bf Find domain:}  \(2ln|1 - t| + C \geq 0\)

\(2ln|1 - t|  \geq -C\)

\(ln|1 - t| \geq -{C \over 2}\)  \hfill Note:  we want to find domain 
for this \(C\) 

 \(|1 - t| \geq e^{-{C \over 2}}\)  since \(e^x\) is an increasing function.

 \(1 - t \leq -e^{-{C \over 2}}\) or  \(1 - t \geq e^{-{C \over 2}}\)

 \(- t \leq -e^{-{C \over 2}} - 1\) or  \(-t \geq e^{-{C \over 2}}- 1\)

 Domain: \(\cases{  t \geq e^{-{C \over 2}} + 1 & if \)t_0 > 0$ \cr
t \leq -e^{-{C \over 2}} + 1 &  if \(t_0 < 0\).}$

Note:  Domain is much easier to determine when the ODE is linear.

 
{\bf Find C given \(y(t_0) = y_0\):}  \(y_0 =  2  \pm  qrt{2ln|1 - t_0| + C}\)

\(\pm(y_0 -2) =     qrt{2ln|1 - t_0| + C}\)

\((y_0 -2)^2 - 2ln|1 - t_0| = C\)

\(y = 2  \pm  qrt{2ln|1 - t| + C}\)

\(y = 2  \pm  qrt{2ln|1 - t| + (y_0 -2)^2 - 2ln|1 - t_0|}\)
 
\(y = 2  \pm  qrt{ (y_0 -2)^2 + ln{(1 - t)^2 \over (1-t_0)^2}}\)

 Domain: \(\cases{  t \geq e^{-{C \over 2}} + 1 & if \)t_0 > 0$ \cr
t \leq -e^{-{C \over 2}} + 1 &  if \(t_0 < 0\).}$

$e^{-{C \over 2}} = e^{-{(y_0 -2)^2 - 2ln|1 - t_0| \over 2}}
= |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} }$

 Domain: \(\cases{  t \geq 1 + |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} } & if \)t_0 > 0$ \cr
t \leq 1 - |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} } &  if \(t_0 < 0\).}$

---
Section 2.5:  

Exponential Growth/Decay \hfill \break
Example:  population growth/radioactive decay)

\(y' = ry\), \(y(0) = y_0\) implies \(y = y_0 e^{rt}\)

\(r > 0\) \hskip 1.2in \(r < 0\)

Logistic growth:  \(y' = h(y)y\)

Example:  \(y' = r(1 - {y \over K})y\)

\(y\) vs \(f(y)\) \hskip 1in slope field:

Equilibrium solutions:

Asymptotically stable:
 fill
Asymptotically unstable:
 fill
Asymptotically semi-stable:

 fill
As \(t \rightarrow \infty\), if \(y > 0\), \(y \rightarrow \)
 fill

Solution:  \(y = {y_0K \over y_0 + (K - y_0)e^{-rt}}\)

{\bf Section 2.7 Euler method:}  Using tangent lines to approximate a function. 

 
\(y_{i+1} = y_i + \Delta y = y_i + {\Delta y \over \Delta t}\Delta t \cong  y_i + {dy \over dt}\Delta t\)

Alternatively use equation of tangent line: 

 slope \(= {y_{i+1 - y_i \over t_{i+1} - t_i} = f'(y_i, t_i)\).}

 
\(y_{i+1} = f'(y_i, t_i)( t_{i+1} - t_i) + y_i = T(t_{i+1})\) where \(y = T(t)\) 

 
Example:  \({dy \over dt} = y^2\), \(y(2) = 1\) implies 
\(y = {1 \over 3 - t}\). 

 box{ \offinterlineskip \halign{  trut 
 rule \hfil  ~# \hfil 
& rule width 1.25 pt \hfil \hskip 0.1in~# \hskip 0.1in\hfil
& rule width 1.25 pt\hfil \hskip 0.2in~#\hskip 0.2in \hfil
 rule \cr
\noalign{---ule}
\(t\) & \(y = 1/(3-t)\)& approximation \cr 
\noalign{---ule height 1.5pt}
2.000000 & 1.000000 & 1.000000 \cr 
 \noalign{ ---ule}
3.000000 & 999.000000 & 2.000000 \cr 
 \noalign{ ---ule}
4.000000 & -1.000000 & 6.000000 \cr 
 \noalign{ ---ule}
5.000000 & -0.500000 & 42.000000 \cr 
 \noalign{ ---ule}
6.000000 & -0.333333 & 1806.000000 \cr 
 \noalign{ ---ule}
 %\noalign{ ---ule}
 %\noalign{ ---ule}
 %\noalign{ ---ule}
 %\noalign{ ---ule}
}
}

 fill

 box{ \offinterlineskip \halign{  trut 
 rule \hfil  ~# \hfil 
& rule width 1.25 pt \hfil \hskip 0.1in~# \hskip 0.1in\hfil
& rule width 1.25 pt\hfil \hskip 0.2in~#\hskip 0.2in \hfil
 rule \cr
\noalign{---ule}
\(t\) & \(y = 1/(3-t)\)& approximation \cr 
\noalign{---ule height 1.5pt}
2.000000 & 1.000000 & 1.000000 \cr 
 \noalign{ ---ule}
2.100000 & 1.111111 & 1.100000 \cr 
 \noalign{ ---ule}
2.200000 & 1.250000 & 1.221000 \cr 
 \noalign{ ---ule}
2.300000 & 1.428571 & 1.370084 \cr 
 \noalign{ ---ule}
2.400000 & 1.666667 & 1.557797 \cr 
 \noalign{ ---ule}
2.500000 & 2.000000 & 1.800470 \cr 
 \noalign{ ---ule}
2.600000 & 2.500000 & 2.124640 \cr 
 \noalign{ ---ule}
2.700000 & 3.333333 & 2.576049 \cr 
 \noalign{ ---ule}
2.800000 & 5.000000 & 3.239652 \cr 
 \noalign{ ---ule}
2.900000 & 10.000004 & 4.289186 \cr 
 \noalign{ ---ule}
}
}

 box{ \offinterlineskip \halign{  trut 
 rule \hfil  ~# \hfil 
& rule width 1.25 pt  \hfil  ~# \hfil 
& rule width 1.25 pt\ \hfil  ~# \hfil 
 rule \cr
\noalign{---ule}
\(t\) & \(y = 1/(3-t)\)& approximation \cr 
\noalign{---ule height 1.5pt}
2.00 & 1.000000 & 1.000000 \cr 
 \noalign{ ---ule}
2.01 & 1.010101 & 1.010000 \cr 
 \noalign{ ---ule}
2.02 & 1.020408 & 1.020201 \cr 
 \noalign{ ---ule}
2.03 & 1.030928 & 1.030609 \cr 
 \noalign{ ---ule}
2.04 & 1.041667 & 1.041231 \cr 
 \noalign{ ---ule}
2.05 & 1.052632 & 1.052072 \cr 
 \noalign{ ---ule}
2.06 & 1.063830 & 1.063141 \cr 
 \noalign{ ---ule}
2.07 & 1.075269 & 1.074443 \cr 
 \noalign{ ---ule}
2.08 & 1.086957 & 1.085988 \cr 
 \noalign{ ---ule}
2.09 & 1.098901 & 1.097782 \cr 
 \noalign{ ---ule}
2.10 & 1.111111 & 1.109833 \cr 
 \noalign{ ---ule}
2.11 & 1.123595 & 1.122150 \cr 
 \noalign{ ---ule}
2.12 & 1.136364 & 1.134742 \cr 
 \noalign{ ---ule}
2.13 & 1.149425 & 1.147619 \cr 
 \noalign{ ---ule}
 \noalign{ ---ule}
 \noalign{ ---ule}
 \noalign{ ---ule}
2.87 & 7.692308 & 6.721314 \cr 
 \noalign{ ---ule}
2.88 & 8.333333 & 7.173075 \cr 
 \noalign{ ---ule}
2.89 & 9.090908 & 7.687605 \cr 
 \noalign{ ---ule}
2.90 & 9.999998 & 8.278598 \cr 
 \noalign{ ---ule}
2.91 & 11.111107 & 8.963949 \cr 
 \noalign{ ---ule}
2.92 & 12.499993 & 9.767473 \cr 
 \noalign{ ---ule}
2.93 & 14.285716 & 10.721509 \cr 
 \noalign{ ---ule}
2.94 & 16.666666 & 11.871017 \cr 
 \noalign{ ---ule}
2.95 & 19.999996 & 13.280227 \cr 
 \noalign{ ---ule}
2.96 & 24.999987 & 15.043871 \cr 
 \noalign{ ---ule}
2.97 & 33.333298 & 17.307051 \cr 
 \noalign{ ---ule}
2.98 & 49.999897 & 20.302391 \cr 
 \noalign{ ---ule}
2.99 & 99.999496 & 24.424261 \cr 
 \noalign{ ---ule}
}
}

\end

$\matrix{ t & y & approximation \cr 
2 & 1 & 1}$

\((1,1/((1-x)(2+x)))/sqrt{1 + 1/((1-x)(2+x))^2\)

\({1,(1-x)(2+x)}/sqrt{1 + ((1-x)(2+x))^2}\)

\({1,(1-y)(2+y)}/sqrt{1 + ((1-y)(2+y))^2}\)

\end

float t;
float y;
float yAprox;
float h = 0.1;

for (i = 0; i < 10; i++ )
{
t = 2 + i*h;
y = 1/(3 - 2 + i*h);
yAprox = yAprox +  yAprox*yAprox*h
printf( ``%d & %d & %d'', t, y, yAprox );
}

START 100/2_8new.tex

2.8:  Approximating solution using 

\bf Method of Successive Approximation
\u
(also called Picard's iteration method).

IVP:   \(y' = f(t, y)\), \(y(t_0) = y_0\).

Note: Can always translate IVP to move initial value to the origin and translate back after solving:

Hence for simplicity {\bf in section 2.8}, we will assume  initial value is at the origin:    \(y' = f(t, y)\), \(y(0) = 0\).

{\bf Thm 2.4.2:}  Suppose the functions \hfill \break 
\(z = f(t, y)\) and \(z = {\partial f \over \partial y}(t, y)\) 
are continuous on \hb
\((a, b) \times (c, d)\) 
and the      
point \((t_0, y_0) \in (a, b) \times (c, d)\), 
then there exists an interval $(t_0 - h, t_0 + h)  ubset 
(a, b)$ such that there exists a unique function 
\(y = \phi(t)\) 
defined on \((t_0 - h, t_0 + h)\) 
that 
satisfies the following
initial value problem:  
\(y'  = f(t, y), ~~y(t_0) = y_0.\)

Thm 2.8.1 is translated to  origin version of Thm 2.4.2:

{\bf Thm 2.8.1:}  Suppose the functions \hfill \break 
\(z = f(t, y)\) and \(z = {\partial f \over \partial y}(t, y)\) 
are continuous for all \(t\) in 
\((-a, a) \times (-c, c)\), 

\u
then there exists an interval $(- h,  h)  ubset 
(-a, a)$ such that there exists a unique function 
\(y = \phi(t)\) 
defined on \((- h, h)\) 
that 
satisfies the following
initial value problem:  
 
\(y'  = f(t, y), ~~y(0) = 0.\)
{\bf Proof outline} (note this is a constructive proof and thus the proof is very useful).

Given:  \(y' = f(t, y)\), \(y(0) = 0\)  \hfil Eqn (*)
\u
\(f\), \(\partial f/\partial y\) continuous \(\forall (t, y) \in (-a, a) \times (-b, b)\).  

Then
\(y = \phi(t)\) is a solution to (*) iff

\(\p'(t) = f(t, \p(t))\), ~~\(\p(0) = 0\) iff

 \(\int_0^t \p'(s) ds = \i f(s, \phi(s)) ds\), ~~\(\p(0) = 0\) iff
 
\( \p(t) = \p(t) - \p(0) =   \i f(s, \phi(s)) ds\) 

Thus \(y = \p(t)\) is a solution to (*) 

Construct \(\phi\) using method of successive approximation --  also called Picard's iteration method.

Let \(\phi_0(t) = 0\) (or the function of your choice)

Let \(\p_1(t) =   \i f(s, \phi_0(s)) ds\) 

Let \(\p_2(t) =   \i f(s, \phi_1(s)) ds\) 

\( dots\)

Let \(\p_{n+1}(t) =   \i f(s, \phi_n(s)) ds\) 

Let \(\phi(t) = \lim_{n \rightarrow \infty} \p_n(t)\)

To finish the proof, need to answer the following questions (see book or more advanced class):

1.)  Does \(\p_n(t)\) exist for all \(n\)?

2.)  Does sequence \(\phi_n\) converge?

3.)  Is \(\phi(t) = \lim_{n \rightarrow \infty} \p_n(t)\) a solution to (*).

4.)  Is the solution unique.

Example:  \(y' = t + 2y\).  \hfil That is \(f(t, y) = t + 2y\)

Let \(\phi_0(t) = 0\)

Let \(\p_1(t) =   \i f(s, 0)ds = \i (s + 2(0))ds\)

Let \(\p_2(t) =   \i f(s, \p_1(s))ds =   \i f(s, {s^2 \over 2})ds\)

Let \(\p_3(t) =   \i f(s, \p_2(s))ds =   \i f(s, {s^2 \over 2}+  {s^3 \over 3})ds\)

Let \(\p_4(t) =   \i f(s, \p_3(s))ds   \)

\hskip 0.59in \(= \i f(s, {s^2 \over 2}+  {s^3 \over 3}+ {s^4 \over 6} )ds\)

\hskip 0.59in \( = \i (s + 2({s^2 \over 2}+  {s^3 \over 3} + {s^4 \over 6}))ds\)

\hskip 0.59in\( =  {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}\)

\( dots\)

Determine formula for \(\p_n\):

Note patterns:

\(\i s ds= {t^2 \over 2} = \)

\(\i {s^2 \over 2}ds = {t^3 \over 3 \cdot 2} = \)

$\i {s^3 \over 3 \cdot 2}ds = {t^4 \over 4 \cdot 
3 \cdot 2} = $

$\i {s^4 \over 4 \cdot 3 \cdot 2}ds = {t^5 \over 5 \cdot 4 \cdot 
3 \cdot 2} = $

Thus look for factorials.

 \(\phi_0(t) = 0\)

 \(\p_1(t) = {t^2 \over 2} \) 

 \(\p_2(t)  =   {t^2 \over 2} +  {t^3 \over 3}\)

 \(\p_3(t) =    {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6}\)

 \(\p_4(t) =   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}=   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 3 \cdot 2} + {t^5 \over 5 \cdot 3}\)

 fill

Thus \(\p_n(t) =\)

FYI (ie not on quizzes/exam):
Defn:  \(\So a_kx ^k = \displaystyle\lim_{n \rightarrow \infty}  o a_kx^k\)

{Taylor's Theorem}:  If \(f\) is analytic  at 0, then for small \(x\) (i.e., \(x\) near 0), 
\(f(x)=\displaystyle um_{k=0^\infty{f^{(k)}(0) \over k!}x^k\) \hskip 1.13in}
 
Example:
 
\(e^t = \So {t^k \over k!}\)  and thus \(e^{bt} = \So {b^kt^k \over k!}\) for \(t\) near 0.

\(\p_n(t) = \displaystyle um_{k = 2}^n {2^{k-2} \over k!} t^k\)
Thus $\p(t) = \displaystyle\lim_{n \rightarrow \infty}\p_n(t) = 
\displaystyle um_{k = 2}^\infty {2^{k-2} \over k!} t^k
 = 
{1 \over 4}\displaystyle um_{k = 2}^\infty {2^{k} \over k!} t^k$
 \rightline{= \( \displaystyle{1 \over 4}\displaystyle\left(~~~~~~~~~~~- ~~~~~~~~~ - ~~~~~~~~~\displaystyle\right)\)}

2.8:  Approximating soln to IVP using seq of fns.

\(\phi_0(t) = 0\), ~~ \textcolor{red}{\(\p_1(t) = {t^2 \over 2} \)},~~
  \textcolor{magenta}{\(\p_2(t)  =   {t^2 \over 2} +  {t^3 \over 3}\)},

 \(\p_4(t) =   {t^2 \over 2} +  {t^3 \over 3} +  {t^4 \over 6} + {t^5 \over 15}\)

2.7: Approximating soln to IVP using multiple tangent lines.

$y(t) = \cases{ 
0 & \(0 \leq t \leq 0.1\) \cr
0.1t -0.01  & \(0.1 \leq t \leq 0.2\) \cr
0.22t -0.034 & \(0.2 \leq t \leq 0.3\) \cr
0.364t -0.0772 & \(0.3 \leq t \leq 0.4\) \cr
0.5328t -0.14672 & \(0.4 \leq t \leq 0.5\) \cr
$}

\end

START 100/FALL18/quiz2_2018ans.tex part 3

[10]~ 2.)  Suppose \({\phi_n}\) is defined by sucessive approximation where  \(y' = {2t \over y+1}\), ~\(y(0) = 0\).
If \(\phi_1(t) = t^2\), then \(\phi_2(t) = \underline{~~ ln|t^2 + 1|~~}\)  \hfill (from 2.8)

\(y' = f(t, y(t))\), \(y(0) = 0\).  Thus \(y(t)  = \int_0^t f(s, y(s))ds\). 

\(\phi_2(t) = \int_0^t f(s, \phi_1(s))ds= \int_0^t {2s \over s^2 + 1} = ln|s^2 + 1| |_0^t  = ln|t^2 + 1| -  ln|1| =  ln|t^2 + 1| \)

 fill
 fill
 fill

\end

START 100/FALL16/e1_Fall2016ANS.tex part 4

3.)  Suppose \(y' = y - t + 1\),     ~\(y(0) = 0\).  

Let \(\phi_0(t) = 0\) and define \(\{\phi_n(t)\}\) by the method of successive approximation (i.e, Picards iteration method).  Determine the following:
 fill

\(y' = f(t, y)\)

\(\phi_1(t) =   \i f(s, \phi_0(s)) ds =  \i f(s, 0) ds =  \i (0 - s + 1)ds =\)

\((- {s^2 \over 2} + s)\e = - {t^2 \over 2} + t - 0\) 
 fill
[3] ~3a) \(\phi_1(t) = \underline{~~- {t^2 \over 2} + t~~}\)
 fill
\(\phi_2(t) =   \i f(s, \phi_1(s)) ds =  \i f(s, - {s^2 \over 2} + s ) ds =  \i (- {s^2 \over 2} + s - s + 1)ds \)

$=  \i (- {s^2 \over 2}  + 1)ds=
(- {s^3 \over 6} + s)\e = - {t^3 \over 6} + t - 0$ 
 fill
[3] ~3b) \(\phi_2(t) = \underline{~~ - {t^3 \over 6} + t~~}\)
 fill
\(\phi_3(t) =   \i f(s, \phi_2(s)) ds =  \i f(s, - {s^3 \over 6} + s ) ds =  \i (- {s^3 \over 6} + s - s + 1)ds \)

$=  \i (- {s^3 \over 6}  + 1)ds=
(- {s^4 \over 24} + s)\e = - {t^3 \over 24} + t - 0$ 

 fill
[3] ~3c) \(\phi_3(t) = \underline{~~- {t^4 \over 24} + t ~~}\)
 fill

 fill
[4] ~3d) \(\phi_n(t) = \underline{~~- {t^{n+1} \over (n+1)!} + t ~~}\)
 fill

 fill
[3] ~3e) \(lim_{n\rightarrow \infty} \phi_n(t) = \underline{~~t~~}\)

[2] ~3f)  Is \(\phi(t) = lim_{n\rightarrow \infty} \phi_n(t)\) a solution to  
\(y' = y - t + 1\),     ~\(y(0) = 0\)?  \(\underline{~~yes~~}\)
 fill

 fill
[2] ~3g)  Is \(\phi(t) = lim_{n\rightarrow \infty} \phi_n(t)\) the unique solution to  
\(y' = y - t + 1\),     ~\(y(0) = 0\)?  \(\underline{~~yes~~}\)

\end

START 100/2_4examples.tex

If possible {\bf without solving}, determine where the solution exists for the following initial value problems:

If not possible  {\bf without solving}, state where in the \(ty\)-plane, the hypothesis of theorem 2.4.2 is satisfied.  In other words, use theorm 2.4.2 to determine where for some interval about \(t_0\), a solution to IVP, \(y' = f(t, y)\), \(y(t_0) = y_0\) exists and is unique. 

Example 1:  \(y' = y^{1 \over 3}, ~y(t_0) = y_0\)
 fill
Example 2:  \(ty' - y = 1, ~y(t_0) = y_0\)
 fill
Example 3:  \(y'  = ln|{t \over y}|, ~y(3) = 6\)
 fill fill
Example 4:  \((t^2 - 1)y' - {t^3y \over t - 4} = ln|t|, ~y(3) = 6\)
 fill

\end

START 100/2_4examples.tex

If possible {\bf without solving}, determine where the solution exists for the following initial value problems:

If not possible  {\bf without solving}, state where in the \(ty\)-plane, the hypothesis of theorem 2.4.2 is satisfied.  In other words, use theorm 2.4.2 to determine where for some interval about \(t_0\), a solution to IVP, \(y' = f(t, y)\), \(y(t_0) = y_0\) exists and is unique. 

Example 1:  \(y' = y^{1 \over 3}, ~y(t_0) = y_0\)
 fill
Example 2:  \(ty' - y = 1, ~y(t_0) = y_0\)
 fill
Example 3:  \(y'  = ln|{t \over y}|, ~y(3) = 6\)
 fill fill
Example 4:  \((t^2 - 1)y' - {t^3y \over t - 4} = ln|t|, ~y(3) = 6\)
 fill

\end

START 100/ch3_134.tex part 1

Derivation of general solutions:

Ch 3: we guessed \(e^{rt}\) is a solution and noted that any linear combination of 
solutions is a solution to a homogeneous linear differential equation.

Section 3.1:   If \(b^2 - 4ac > 0\):  ~~
\(y = c_1e^{r_1t} + c_2e^{r_2t}\)

START 100/ch3_134.tex AND 100/ch3new.tex

{\bf Examples:}  
\u\u
Ex 1:  Solve \(y''- 3y' - 4y = 0\), ~\(y(0) = 1\), \(y'(0) = 2\).

If \(y = e^{rt\), then \(y' = re^{rt}\) and \(y'' = r^2e^{rt}\).}
\u
\(r^2e^{rt} - 3re^{rt} - 4e^{rt} = 0\) 
\u
\(r^2 - 3r - 4 = 0\) implies \((r -4)(r+1) = 0\).  Thus \(r = 4, -1\)
\u

Hence general solution is \(y = c_1e^{4t  + c_2e^{-t}\)}
\u
Solution to IVP:  
\u\u\u
Need to solve for 2 unknowns, \(c_1\) \& \(c_2\); thus need 2 eqns:
\u
 \(y = c_1e^{4t}  + c_2e^{-t}\), ~~~~\(y(0) = 1\) ~~implies~~ \(1 = c_1 + c_2\)
 \u
\(y' = 4c_1e^{4t}  - c_2e^{-t}\), ~~ \(y'(0) = 2\)~~implies~~\(2 = 4c_1  - c_2\) 
\u
Thus \(3 = 5c_1\) \& hence \(c_1 = {3 \over 5}\) and \(c_2 = 1 - c_1 = 1 -  {3 \over 5} =  {2 \over 5}\)

Thus IVP soln:   \(y =  {3 \over 5e^{4t}  +  {2\over 5}e^{-t}\)}

---
\u

Ex:  linear homogeneous 1rst order DE: \(y' + 2y = 0\)

integrating factor \(u(t) = e^{\int 2dt} = e^{2t}\)

\(y'e^{2t} + 2e^{2t}y = 0\)

\((e^{2t}y)' = 0\).  Thus \(\int (e^{2t}y)'dt = \int 0 dt\).  Hence \(e^{2t}y = C\)

So \(y = Ce^{-2t}\).  

Thus exponential function could also be a solution to a linear homogeneous 2nd order DE

Ex: Simple linear homog 2nd order DE  \(y'' +2 y' = 0\).

Let \(v = y'\), then \(v' = y''\)

 \(y'' +2 y' = 0\) implies  \(v' +2 v = 0\) implies \(v = e^{2t}\).

Thus \(v = y' = {dy \over dt} = Ce^{-2t}\).  Hence \(dy = Ce^{-2t}dt\) and
 fill
 \(y = c_1e^{-2t + c_2\).}
 \(y = c_1e^{-2t + c_2\).}

Note 2 integrations give us 2 constants.

Note also that we the general solution is a linear combination of two solutions:

Let \(c_1 = 1, ~c_2 = 0\), then we see, \(y(t) = e^{-2t}\) is a solution.

 
Let \(c_1 = 0, ~c_2 = 1\), then we see, \(y(t) = 1\) is a solution.

The general solution is a linear combination of two solutions:
\u
 \(y = c_1e^{-2t + c_2(1)\).}

---

Recall:  you have seen this before:  
\u
Solve linear homogeneous matrix equation \(A{\bf y} = {\bf 0}\).  
\u
The general solution is a linear combination of linearly independent vectors that span the solution space:  

\({\bf y = c_1{\bf v_1} + ... c_n{\bf v_n}\)}

---

\u

{FYI:  You could see this again:
\u
Math 4050:  Solve homogeneous linear recurrance relation \(x_n - x_{n-1} - x_{n-2}=0\)  where \(x_1 = 1\) and \(x_2 = 1\).
\u
Fibonacci sequence:  \(x_n = x_{n-1} + x_{n-2}\)
1, 1, 2, 3, 5, 8, 13, 21, ...
\u
Note $x_n ={1 \over  qrt{5}}({1 +  qrt{5} \over 2})^n -
{1 \over  qrt{5}}({1 -  qrt{5} \over 2})^n$

Proof: \(x_n = x_{n-1} + x_{n-2}\)  implies \(x_n - x_{n-1} - x_{n-2} = 0\)

  Suppose \(x_n = r^n\).  Then \(x_{n-1} = r^{n-1}\) and \(x_{n-2} = r^{n-2}\)

Then 
\(0 = x_n - x_{n-1} - x_{n-2} = r^n -  r^{n-1} -  r^{n-2}\)

Thus \(r^{n-2}(r^2 - r - 1) = 0\).  

Thus either \(r = 0\) or $r = {1 \pm  qrt{1 - 4(1)(-1)} \over 2} = 
{1 \pm  qrt{5} \over 2}$ 

Thus \(x_n = 0\), ~~\(x_n = \left({1 +  qrt{5} \over 2}\right)^n\) and 
~~\(f_n = \left({1 -  qrt{5} \over 2}\right)^n\) 
are
3 different sequences that satisfy the
\u
homog linear recurrence relation: \(x_n - x_{n-1} - x_{n-2} = 0\).

Hence $x_n = c_1\left({1 +  qrt{5} \over 2}\right)^n + c_2 
\left({1 -  qrt{5} \over 2}\right)^n$ 
{also satisfies this}

homogeneous linear recurrence relation.

Suppose the initial conditions are \(x_1 = 1\) and \(x_2 = 1\)

Then for \(n = 1\):  \(x_1 = 1\) implies \( c_1 + c_2 = 1\)

For \(n = 2\):  \(x_2 = 1\) implies  $c_1\left({1 +  qrt{5} \over 2}\right) + c_2 
\left({1 -  qrt{5} \over 2}\right) = 1$}

We can solve this for \(c_1\) and \(c_2\) to determine that 
   \t
 $x_n ={1 \over  qrt{5}({1 +  qrt{5} \over 2})^n -
{1 \over  qrt{5}}({1 -  qrt{5} \over 2})^n$}

{\bf Second order differential equation:}
 
Linear equation with constant coefficients:
\hskip 10pt
If the second order differential equation is 
\(ay'' + by' + cy = 0\), 

then \(y = e^{rt\) is a solution}

Need to have two independent solutions.

Solve the following IVPs:

1.)  \(y'' - 6y' + 9y = 0\)    \hfill     \(y(0) = 1, ~ y'(0) = 2\)

\w

2.) \(4y'' - y' + 2y = 0\) \hfill     \(y(0) = 3, ~ y'(0) = 4\)
\w

3.) \(4y'' + 4y' + y = 0\)  \hfill     \(y(0) = 6, ~ y'(0) = 7\)
\w

4.)  \(2y'' - 2y = 0\)  \hfill     \(y(0) = 5, ~y'(0) = 9\)

Thm:  Suppose \(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

If \(\psi\) is a solution to

\(ay'' + by' + cy = g(t)\) [*],
Then \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is also a solution to [*].

Moreover if \(\gamma\) is also a solution to [*], then there exist constants \(c_1, c_2\) such that 

\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\)

Or in other words, \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to [*].

Proof:  

Define   \(L(f) = af'' + bf' + cf\).

Recall \(L\) is a linear function.

 Let \(h = c_1 \phi_1(t) + c_2 \phi_2(t)\).  Since \(h\) is a solution to the differential 
equation, \(ay'' + by' + cy = 0\),

Since \(\psi\) is a solution to 
 \(ay'' + by' + cy = g(t)\),

We will now show that \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h\) is also a solution to 
[*].

Since \(\gamma\) a solution to  \(ay'' + by' + cy = g(t)\),

We will first show that \(\gamma - \psi\) is a solution to the differential equation
 \(ay'' + by' + cy = 0\).

 fill

Since \(\gamma - \psi\) is a solution to \(ay'' + by' + cy = 0\) and 

\(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

there exist constants \(c_1, c_2\) such that

\(\gamma - \psi =\underline{\hskip 2in\)}

Thus
\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\).

Thm: \hfil \break
 Suppose  
\(f_1\) is a a solution to
\(ay'' + by' + cy = g_1(t)\)  \hfil \break
and \(f_2\) is a a solution to
\(ay'' + by' + cy = g_2(t)\), then \(f_1 + f_2\) is a solution to
\(ay'' + by' + cy = g_1(t) + g_2(t)\)

Proof: Let  \(L(f) = af'' + bf' + cf\).

Since \(f_1\) is a solution to 
 \(ay'' + by' + cy = g_1(t)\),

Since \(f_2\) is a solution to 
 \(ay'' + by' + cy = g_2(t)\),

We will now show that \(f_1 + f_2\) is a solution to \break
 \(ay'' + by' + cy = g_1(t) + g_2(t)\).

 fill

Sidenote:  The proofs above work even if \(a, b, c\) are functions of \(t\) instead of constants.

\end

START 100/ch3new.tex part 1

Initial value problem:  use \(y(t_0) = y_0\), \(y'(t_0) = y_0'\) to solve for \(c_1, c_2\) to find unique 
solution. 
Derivation of general solutions:

If \(b^2 - 4ac > 0\) we guessed \(e^{rt}\) is a solution and noted that any linear combination of 
solutions is a solution to a homogeneous linear differential equation.

\end

START 100/exam1review.tex part 5

Note:  You must be able to identify which techniques you need to use.  For example:

Integration:

* Integration by substitution

 * Integration by parts

* Integration by partial fractions

Note: Partial fractions are also used in ch 6 for a different application.

For differential equations:

Is the differential equation 1rst order or 2nd order?

If 2nd order:  Section 3.1, solve \(ay'' + by' + cy = 0\).  

Guess \(y = e^{rt}\).

  skip -8pt
\(ar^2e^{rt} + bre^{rt} + ce^{rt} = 0\) implies \(ar^2 + br + c = 0\),
 

Need to have two independent solutions.

If \(y = \phi_1, y = \phi_2\) are solutions to a  LINEAR HOMOGENEOUS differential equation, \(y = c_1\phi_1 + c_2\phi_2\) 
is also 
a solution 

If 1st order:  Is the equation linear or separable or ?

\end

START 100/FALL17/quiz3Fall2017formAans.tex part 2

[2]~ 6.)  Note this is a 3.1 problem.  The general solution to \(y'' + 2y' - 8y = 0\) is
B.)  \(y = c_1e^{2t} + c_2e^{-4t}\)\hfill
~~~~~
 since \(r^2 + 2r - 8 = (r +4)(r-2) = 0\) implies \(r = -4, 2\).

\end

START 100/FALL17/quiz3Fall2017formBans.tex part 2

[2]~ 6.)  Note this is a 3.1 problem.  The general solution to \(y'' - 2y' - 8y = 0\) is
C.)  \(y = c_1e^{-2t} + c_2e^{ 4t}\)\hfill
~~~~~
 since \(r^2 - 2r - 8 = (r -4)(r+2) = 0\) implies \(r = 4, -2\).

\end

START 100/FALL17/quiz4Fall2017formBans.tex part 1

[2]~ 6.)  Note this is a 3.1 problem.  The general solution to \(y'' + 2y' -8y = 0\) is

A.)  \(y = c_1e^{2t} + c_2e^{4t}\) \hfill
B.)  \(y = c_1e^{2t} + c_2e^{-4t}\) \hfill
C.)  \(y = c_1e^{-2t} + c_2e^{4t}\) \hfill

D.)  \(y = c_1e^{-2t} + c_2e^{-4t}\) \hfill
E.)  \(y = c_1e^{2t} + c_2te^{2t}\) \hfill
F.)  \(y = c_1cos(2t) + c_2sin(2t)\)\hfill

\end

START 100/FALL16/e1_Fall2016ANS.tex part 5

[20]~ 1.)  Solve \(y'' - 6y' + 9 = 0\), \(y(0) = 2\), \(y'(0) = 4\).

 fill
Answer: \(\underline{~~\hskip 4in~~\)}

 2.)  Circle T for true and F for false.  

[4]~ 2a.)  The equation \(ln(t)y' = {t \over t+1} -  y(sin t^2)\) is a linear differential equation.

[4]~ 2b.)  The equation \(y' + y = y^2\) is a linear differential equation.
\hfill{T~~~~~~~~~~~F}

[4]~ 2c.)  Suppose \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to 
 \(ay'' + by' + cy = 0\).
If \(y = h(t)\) is   also a solution to  \(ay'' + by' + cy = 0\), then there exists constants \(c_1\) and \(c_2\) such that \(h(t) = c_1\phi_1(t) + c_2\phi_2(t)\).
\hfill{T~~~~~~~~~~~F}

[4]~ 2d.)   Suppose \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are linearly independent solutions to 
 \(ay'' + by' + cy = 0\).
If \(y = h(t)\) is   also a solution to  \(ay'' + by' + cy = 0\), then there exists constants \(c_1\) and \(c_2\) such that \(h(t) = c_1\phi_1(t) + c_2\phi_2(t)\).
\hfill{T~~~~~~~~~~~F}

\end

START 100/SPRING15/e1_2016ans.tex part 3

\nopagenumbers

Math 3600 Differential Equations Exam \#1
                               March 2, 2016 \hfill  SHOW ALL
WORK
~~~

[20]~ 1.)  Solve \(y'' - 6y' + 9y = 0\), \(y(0) = 2\), \(y'(0) = 4\).

\(r^2 - 6r + 9 = (r-3)^2 = 0\).  Thus \(r = 3\)

General solution:  \(y = c_1e^{3t} + c_2te^{3t}\)

 \(y' = 3c_1e^{3t} + c_2(e^{3t} + 3te^{3t})\)

 \(y(0) = 2\):  ~~~~\(2 = c_1\)
 
 \(y'(0) = 4\): ~~~~\(4 = 3c_1 + c_2\).  Thus \(c_2 = 4 - 6 = -2\).

 fill
Answer: \(\underline{~~y = 2e^{3t - 2te^{3t}~~}\)}

\end

START 34/FALL03/finalexamANS.txt part 4

\nopagenumbers

Math 34 Differential Equations Final Exam
                               December 15, 2003 \hfill  SHOW ALL
WORK
~

[3]~ 1a.)  Without using the LaPlace transform, solve the following initial value problem:  

\(y'' + 2y' + y = 0, ~y(0) = 0, ~y'(0) = 4\)

Suppose \(y = e^{rt}\).  Then \(y' = re^{rt}, ~y'' = r^2e^{rt}\)

\(r^2e^{rt} + 2re^{rt} + e^{rt} = 0\)
Hence \(r^2 + 2r + 1 = 0\).  Thus, \((r + 1)^2 = 0\).  Hence \(r = -1\)

Hence general solution is \(y(t) = c_1e^{-t} + c_2te^{-t}\)

\(y(0) = 0:  0 = c_1 + 0\).  Thus \(c_1= 0\)

\(y'(0) = 4:  y' = c_2(e^{-t} -te^{-t})\).  Thus \(4 = c_2(1 - 0) = c_2\)
  

Thus, \(y(t) = 4te^{-t}\) is the solution to the initial value problem.
\u
Answer 1a.) \(\underline{4te^{-t}\)}
\w\u

\end

START 100/3_2.tex

 
{\bf Existence and Uniqueness for  LINEAR DEs.}
\u\u
\underbar{Homogeneous:}
 

\(y^{(n) + p_1(t)y^{(n-1)} + ... p_{n-1}(t) y' + p_n(t)y = 0\)}
\u
\underbar{Non-homogeneous:} \(g(t) \not= 0\)
 

\(y^{(n) + p_1(t)y^{(n-1)} + ... p_{n-1}(t) y' + p_n(t)y = g(t)\)}
\u
{\bf 1st order LINEAR differential equation:}
\u\u

Thm 2.4.1:  If \(p:(a, b) \rightarrow R\) and \(g:(a, b) \rightarrow R\) are continuous and $a < 
t_0 < b$, then 
there exists a unique function \(y = \phi(t)\), ~\(\phi:(a, b) \rightarrow R\) that satisfies the 
 
IVP:  \(y' + p(t) y = g(t)\), ~~\(y(t_0) = y_0\)

\u
Thm:  If \(y = \phi_1(t)\) is a solution to \underbar{homogeneous} equation, \(y' + p(t) y = 0\), then \(y = c\phi_1(t)\) is the general solution to this equation.
\u\u
If in addition \(y = \psi(t)\) is a solution to \underbar{non-homogeneous} equation, \(y' + p(t) y = g(t)\), then \(y = c\phi_1(t) + \psi(t)\) is the general solution to this equation.

Partial proof:  \(y = \phi_1(t)\) is a solution to  \(y' + p(t) y = 0\)
implies

Thus \(y = c\phi_1(t)\) is a solution to  \(y' + p(t) y = 0\) since
 fill

 \(y = \psi(t)\) is a solution to  \(y' + p(t) y = g(t)\)
implies
 fill
Thus \(y = c\phi_1(t) + \psi(t)\) is a solution to  \(y' + p(t) y = g(t)\) since

{\bf 2nd order LINEAR differential equation:}
\u

Thm 3.2.1:  If \(p:(a, b) \rightarrow R\), \(q:(a, b) \rightarrow R\), and \(g:(a, b) \rightarrow R\) 
are 
continuous and \(a < t_0 < b\), then
there exists a unique function \(y = \phi(t)\), \(\phi:(a, b) \rightarrow R\) that satisfies the
initial value problem

\(y'' + p(t) y' + q(t)y = g(t)\),

\(y(t_0) = y_0\),

\(y'(t_0) = y_0'\)

Thm 3.2.2:  If \(\phi_1\) and \(\phi_2\) are two solutions to a \underbar{homogeneous} linear differential 
equation, then 
\(c_1\phi_1 + c_2\phi_2\) is also a solution to this linear differential equation.

Proof of thm 3.2.2:  

Since \(y(t) = \phi_i(t)\) is a solution to the linear homogeneous differential equation \(y'' + py' + qy = 0\) where \(p\) and \(q\) are functions of \(t\) (note this includes the case with constant coefficients), then

Claim:   \(y(t) = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution to \(y'' + py' + qy = 0\) 

Pf of claim:  

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).
 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\).
 
Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

{\bf When does the following IVP have  unique sol'n:}

IVP:  \(ay'' + by' + cy = 0\), \(y(t_0) = y_0\), \(y'(t_0) = y_1\).

Suppose \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is a solution to 
 
\(y' = c_1\phi_1'(t) + c_2\phi_2'(t)\)}

\(y(t_0) = y_0\):  ~~\(y_0 = c_1\phi_1(t_0) + c_2\phi_2(t_0)\)

\(y'(t_0) = y_1\):  ~~\(y_1 = c_1\phi_1'(t_0) + c_2\phi_2'(t_0)\)

To find IVP solution, need to solve above system of two equations for the unknowns \(c_1\) and \(c_2\).

Note the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\).

Note that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0,  \phi_1(t_0), \phi_2(t_0)\), \(y_1, \phi_1'(t_0), \phi_2'(t_0)\) are the constants.  We can translate this linear system of equations into matrix form:
\u
$\matrix{c_1\phi_1(t_0) + c_2\phi_2(t_0) = y_0 \cr
c_1\phi_1'(t_0) + c_2\phi_2'(t_0) = y_1}\( \hfill \)\Rightarrow$ \hfill
$\left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right]\left[\matrix{c_1 \cr c_2} \right]=\left[\matrix{y_0 \cr y_1} \right] $

Note this equation has a unique solution if and only if 
 
$det \left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0) \right] = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|  = \phi_1 \phi_2' - \phi_1'\phi_2 \not= 0$ }

Definition:  The Wronskian of two differential functions, \(\phi_1\) and \(\phi_2\) is 
\u\u
\(W(\phi_1, \phi_2) = \phi_1 \phi_2' - \phi_1'\phi_2 = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'\right|\)}

Examples:
\u\u\u
1.)  W(\(cos(t)\), \(sin(t)\)) =
\(\left|\matrix{cos(t) & sin(t) \cr -sin(t) & cos(t)}\right| \)
  
\(~~~~~~~~~~~~~~~~= cos^2(t) + sin^2(t) = 1 > 0.\) 

2.)  W(\(e^{dt}cos(nt)\), \(e^{dt}sin(nt)) =\) 
  
~~~~$ \left|\matrix{e^{dtcos(nt) && e^{dt}sin(nt) \cr
 de^{dt}cos(nt)- ne^{dt}sin(nt) &~& de^{dt}sin(nt) + ne^{dt}cos(nt)}\right|$}

\(^{= e^{dt}cos(nt)(de^{dt}sin(nt) + ne^{dt}cos(nt) ) - e^{dt}sin(nt)( de^{dt}cos(nt)- ne^{dt}sin(nt) )}\)

\u\u
$~^{= e^{2dt}%\huge
[cos(nt)(dsin(nt) + ncos(nt)) - sin(nt)(dcos(nt)- nsin(nt) )]%\huge
}$
\u\u
$~^{= e^{2dt}%\huge
[dcos(nt)sin(nt) + ncos^2(nt) - dsin(nt)cos(nt)+ nsin^2(nt)]%\huge
)}$
\u\u

$~~= e^{2dt}%\huge
[ ncos^2(nt) + nsin^2(nt)%\huge
]$
  
[ cos^2(nt) + sin^2(nt)%\huge
]$
\(~=~ ne^{2dt} > 0\) for all \(t\).}

Definition:  The Wronskian of two differential functions, \(f\) and \(g\) is 

\(W(f, g) = fg' - f'g = \big|\matrix{f & g \cr f' & g'\big|\)}

Thm 3.2.3:  Suppose that 
 \(\phi_1\) and \(\phi_2\) are two solutions to $y'' + p(t) y' + q(t)y = 
0$. \hb
 There is a unique choice of constants \(c_1\) and \(c_2\) such that \(c_1\phi_1 + c_2\phi_2\) 
satisfies this 
homog linear differential 
equation and initial conditions, \(y(t_0) = y_0\), \(y'(t_0) = y_0'\).

iff
  \(W(\phi_1, \phi_2)(t_0) = \phi_1(t_0)\phi_2'(t_0) - \phi_1'(t_0)\phi_2(t_0) \not= 0\).

 fill
Thm 3.2.4:  Given the hypothesis of thm 3.2.1, \hb
suppose that \(\phi_1\) and \(\phi_2\) are two solutions to 
 
\(y'' + p(t) y' + q(t)y =0\).
 \u\u
 If \(W(\phi_1, \phi_2)(t_0) \not= 0\), 
for some \(t_0 \in (a, b)\), then any solution to this homogeneous linear differential
equation can be written as $y = 
c_1\phi_1 + c_2\phi_2\( for some \)c_1\( and \)c_2$.
 fill
Defn  If \(\phi_1\) and \(\phi_2\) satisfy the conditions in thm 3.2.4, then \(\phi_1\) and \(\phi_2\) 
form a fundamental set of solutions to $y'' + p(t) y' + q(t)y =
0$. 

 fill
Thm 3.2.5:  Given any second order homogeneous linear differential equation, there exist a pair 
of functions which form a fundamental set of solutions.

FYI:  Linear Independence and the Wronskian

Defn: \(\po\) and \(\pt\) are linearly dependent if there exists constants \(c_1, c_2\) 
such that \(c_1 \not= 0\) or \(c_2 \not= 0\) and 

\(c_1\po(t) + c_2\pt(t) = 0\) for all \(t \in (a, b)\)

Thm 3.3.1:  If \(\po: (a, b) \rightarrow R\) and \(\pt(a, b) \rightarrow R\) are differentiable 
functions on (a, b) and \hb if \(W(\po, \pt)(t_0) \not= 0\) for some \(t_0 \in (a, b)\), then\hb \(\po\) and \(\pt\) 
are linearly independent on \((a, b)\).  \hb Moreover, if \(\po\) and \(\pt\) are linearly dependent on $(a, 
b)\(, then \)W(\po, \pt)(t) = 0\(  for all \)t \in (a, b)$

Proof idea:

If \(c_1\po(t) + c_2\pt(t) = 0\) for all \(t\in (a, b)\),

\(c_1\po'(t) + c_2\pt'(t) = 0\) for all \(t\in (a, b)\)}

Solve the following linear system of equations for \(c_1, c_2\)

$\matrix{c_1 \po(t_0) + c_2\pt(t_0) = 0
\cr
c_1 \po'(t_0) + c_2\pt'(t_0) = 0}$

$\left[\matrix{\po(t_0) & \pt(t_0) \cr
\po'(t_0) & \pt'(t_0)}\right]
\left[\matrix{
c_1 \cr c_2
}\right]
=
\left[\matrix{
0 \cr 0
}\right]$
 fill
In other words the fundamental set of solutions \(\{\po, \pt\}\) to $y'' + p(t) y' + q(t)y =
0$ form a basis for the set of all solutions to this linear homogeneous DE.

\end

Thm:  Suppose \(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

If \(\psi\) is a solution to

\(ay'' + by' + cy = g(t)\) [*],
Then \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is also a solution to [*].

Moreover if \(\gamma\) is also a solution to [*], then there exist constants \(c_1, c_2\) such that 

\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\)

Or in other words, \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to [*].

Proof:  

Define   \(L(f) = af'' + bf' + cf\).

Recall \(L\) is a linear function.

 Let \(h = c_1 \phi_1(t) + c_2 \phi_2(t)\).  Since \(h\) is a solution to the differential 
equation, \(ay'' + by' + cy = 0\),

Since \(\psi\) is a solution to 
 \(ay'' + by' + cy = g(t)\),

We will now show that \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h\) is also a solution to 
[*].

Since \(\gamma\) a solution to  \(ay'' + by' + cy = g(t)\),

We will first show that \(\gamma - \psi\) is a solution to the differential equation
 \(ay'' + by' + cy = 0\).

 fill

Since \(\gamma - \psi\) is a solution to \(ay'' + by' + cy = 0\) and 

\(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

there exist constants \(c_1, c_2\) such that

\(\gamma - \psi =\underline{\hskip 2in\)}

Thus
\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\).

Thm: \hfil \break
 Suppose  
\(f_1\) is a a solution to
\(ay'' + by' + cy = g_1(t)\)  \hfil \break
and \(f_2\) is a a solution to
\(ay'' + by' + cy = g_2(t)\), then \(f_1 + f_2\) is a solution to
\(ay'' + by' + cy = g_1(t) + g_2(t)\)

Proof: Let  \(L(f) = af'' + bf' + cf\).

Since \(f_1\) is a solution to 
 \(ay'' + by' + cy = g_1(t)\),

Since \(f_2\) is a solution to 
 \(ay'' + by' + cy = g_2(t)\),

We will now show that \(f_1 + f_2\) is a solution to \break
 \(ay'' + by' + cy = g_1(t) + g_2(t)\).

 fill

Sidenote:  The proofs above work even if \(a, b, c\) are functions of \(t\) instead of constants.

{\bf Examples:} Find a suitable form for \(\psi\) for the following differential equations:

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 fill

2.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 fill

3.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 fill

4.)  \(y'' - 5y  = 4sin(3t)\)
 fill

5.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 fill

6.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 fill

7.)  
\(y'' - 4y' - 5y = 4 sin(3t) e^{2t}\)
 fill
8.)  
\(y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}\)
 fill

9.) \(y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}\)
 fill
10.)  \(y'' - 4y' - 5y\)

 fill
11.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 fill
12.) \(y'' - 4y' - 5y = 4e^{-t}\)
 fill
To solve \(ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)\) [**]

1.)  Find the general solution to \(ay'' + by' + cy = 0\):  

\(c_1\phi_1 + c_2\phi_2\)

2.)  For each \(g_i\), find a solution to \(ay'' + by' + cy = g_i\):  

\(\psi_i\)

This includes plugging guessed solution {\(\psi_i\)} into \break
\(ay'' + by' + cy = g_i\).

The general solution to [**] is

\(c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n\)

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.) {\bf First solve homogeneous equation:}

  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to \hb
  LINEAR HOMOGENEOUS equation.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.) {\bf Find one solution to non-homogeneous eq'n:}
  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)
~~~~~~~~\(y' = 3A cos(3t) - 3B sin(3t)\)
~~~~~~~~\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.) {\bf If initial value problem:}
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

\((e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0\) and 
\((e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0\) 

Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/3_2a.tex

\documentclass[12pt]{article}

 etlength{\topmargin}{-0.7in}
 etlength{\oddsidemargin}{-.250in}
 etlength{\textwidth}{7.4in}
 etlength{\textheight}{9.6in}
\pagestyle{empty}  %% To avoid page numbering
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{relsize}

\AppendGraphicsExtensions{.gif}
\DeclareGraphicsExtensions{.pdf,.png,.gif,.jpg}
\begin{document}

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).
 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\).
Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

{\bf When does the following IVP have a unique solution:}

IVP:  \(ay'' + by' + cy = 0\), \(y(t_0) = y_0\), \(y'(t_0) = y_1\).

Suppose \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is a solution to \(ay'' + by' + cy = 0\).  Then 
\(y' = c_1\phi_1'(t) + c_2\phi_2'(t)\)

\(y(t_0) = y_0\):  \(y_0 = c_1\phi_1(t_0) + c_2\phi_2(t_0)\)

\(y'(t_0) = y_1\):  \(y_1 = c_1\phi_1'(t_0) + c_2\phi_2'(t_0)\)

To find IVP solution, need to solve above system of two equations for the unknowns \(c_1\) and \(c_2\).

Note the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\).

Note that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0,  \phi_1(t_0), \phi_2(t_0)\), \(y_1, \phi_1'(t_0), \phi_2'(t_0)\) are the constants.  We can translate this linear system of equations into matrix form:

$\matrix{c_1\phi_1(t_0) + c_2\phi_2(t_0) = y_0 \cr
c_1\phi_1'(t_0) + c_2\phi_2'(t_0) = y_1}$ ~~~implies~~~
$\left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right]
\left[\matrix{c_1 \cr c_2} \right] = \left[\matrix{y_0 \cr y_1} \right] $

Note this equation has a unique solution if and only if $det \left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right] = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|  = \phi_1 \phi_2' - \phi_1'\phi_2 \not= 0$ 

Definition:  The Wronskian of two differential functions, \(\phi_1\) and \(\phi_2\) is 

\(W(\phi_1, \phi_2) = \phi_1 \phi_2' - \phi_1'\phi_2 = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'\right|\)}

Examples:

1.)  Wronskian of \(cos(t)\), \(sin(t)\) =
\(\left|\matrix{cos(t) & sin(t) \cr -sin(t) & cos(t)}\right| = cos^2(t) + sin^2(t) = 1 > 0.\) 

2.)  Wronskian of \(e^{dt}cos(nt)\), \(e^{dt}sin(nt)\) =
$\left|\matrix{e^{dt}cos(nt) && e^{dt}sin(nt) \cr
 de^{dt}cos(nt)- ne^{dt}sin(nt) &~& de^{dt}sin(nt) + ne^{dt}cos(nt)}\right|$}

$= e^{dt}cos(nt)[de^{dt}sin(nt) + ne^{dt}cos(nt)] - 
e^{dt}sin(nt)[ de^{dt}cos(nt)- ne^{dt}sin(nt)]$

\(= e^{2dt}\huge(cos(nt)[dsin(nt) + ncos(nt)] - sin(nt)[ dcos(nt)- nsin(nt)]\huge)\)

\(= e^{2dt}\huge(dcos(nt)sin(nt) + ncos^2(nt)] - dsin(nt)cos(nt)+ nsin^2(nt)]\huge)\)

\(= e^{2dt}\huge( ncos^2(nt) + nsin^2(nt)\huge)\)
\(= ne^{2dt}\huge( cos^2(nt) + sin^2(nt)\huge)\)
\(= ne^{2dt} > 0\) for all \(t\).
\end{document}

START 100/ch3.tex part 1

{\bf Existence and Uniqueness for  LINEAR DEs.}
\u\u
\underbar{Homogeneous:}
 

\(y^{(n) + p_1(t)y^{(n-1)} + ... p_{n-1}(t) y' + p_n(t)y = 0\)}
\u
\underbar{Non-homogeneous:} \(g(t) \not= 0\)
 

\(y^{(n) + p_1(t)y^{(n-1)} + ... p_{n-1}(t) y' + p_n(t)y = g(t)\)}
\u
{\bf 1st order LINEAR differential equation:}
\u\u

Thm 2.4.1:  If \(p:(a, b) \rightarrow R\) and \(g:(a, b) \rightarrow R\) are continuous and $a < 
t_0 < b$, then 
there exists a unique function \(y = \phi(t)\), ~\(\phi:(a, b) \rightarrow R\) that satisfies the 
 
IVP:  \(y' + p(t) y = g(t)\), ~~\(y(t_0) = y_0\)

\u
Thm:  If \(y = \phi_1(t)\) is a solution to \underbar{homogeneous} equation, \(y' + p(t) y = 0\), then \(y = c\phi_1(t)\) is the general solution to this equation.
\u\u
If in addition \(y = \psi(t)\) is a solution to \underbar{non-homogeneous} equation, \(y' + p(t) y = g(t)\), then \(y = c\phi_1(t) + \psi(t)\) is the general solution to this equation.

Partial proof:  \(y = \phi_1(t)\) is a solution to  \(y' + p(t) y = 0\)
implies

Thus \(y = c\phi_1(t)\) is a solution to  \(y' + p(t) y = 0\) since
 fill

 \(y = \psi(t)\) is a solution to  \(y' + p(t) y = g(t)\)
implies
 fill
Thus \(y = c\phi_1(t) + \psi(t)\) is a solution to  \(y' + p(t) y = g(t)\) since

{\bf 2nd order LINEAR differential equation:}
\u

Thm 3.2.1:  If \(p:(a, b) \rightarrow R\), \(q:(a, b) \rightarrow R\), and \(g:(a, b) \rightarrow R\) 
are 
continuous and \(a < t_0 < b\), then
there exists a unique function \(y = \phi(t)\), \(\phi:(a, b) \rightarrow R\) that satisfies the
initial value problem

\(y'' + p(t) y' + q(t)y = g(t)\),

\(y(t_0) = y_0\),

\(y'(t_0) = y_0'\)

Thm 3.2.2:  If \(\phi_1\) and \(\phi_2\) are two solutions to a \underbar{homogeneous} linear differential 
equation, then 
\(c_1\phi_1 + c_2\phi_2\) is also a solution to this linear differential equation.

Proof of thm 3.2.2:  

Since \(y(t) = \phi_i(t)\) is a solution to the linear homogeneous differential equation \(y'' + py' + qy = 0\) where \(p\) and \(q\) are functions of \(t\) (note this includes the case with constant coefficients), then

Claim:   \(y(t) = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution to \(y'' + py' + qy = 0\) 

Pf of claim:  

{\bf Second order differential equation:}
 
Linear equation with constant coefficients:
\hskip 10pt
If the second order differential equation is 
\(ay'' + by' + cy = 0\), 

then \(y = e^{rt\) is a solution}

Need to have two independent solutions.

Solve the following IVPs:

1.)  \(y'' - 6y' + 9y = 0\)    \hfill     \(y(0) = 1, ~ y'(0) = 2\)

\w

2.) \(4y'' - y' + 2y = 0\) \hfill     \(y(0) = 3, ~ y'(0) = 4\)
\w

3.) \(4y'' + 4y' + y = 0\)  \hfill     \(y(0) = 6, ~ y'(0) = 7\)
\w

4.)  \(2y'' - 2y = 0\)  \hfill     \(y(0) = 5, ~y'(0) = 9\)

\(ay'' + by' + cy = 0\), ~~\(y = e^{rt}\), then 
\(ar^2e^{rt} + bre^{rt} + ce^{rt} = 0\) implies \(ar^2 + br + c = 0\),
 
Suppose \(r = r_1, r_2\) are solutions to \(ar^2 + br + c = 0\)

\(r_1, r_2 = {-b \pm  qrt{b^2 - 4ac \over 2a}\)}

If \(r_1 \not= r_2\), then \(b^2 - 4ac \not= 0\).  Hence a general solution is
{\(y = c_1e^{r_1t} + c_2e^{r_2t}\)}

If \(b^2 - 4ac > 0\), general solution is \(y = c_1e^{r_1t} + c_2e^{r_2t}\).

If \(b^2 - 4ac < 0\), change format to linear combination of real-valued functions instead of 
complex valued functions by using Euler's formula.

general solution is \(y = c_1 e^{dt} cos (nt) + c_2 e^{dt} sin (nt)\) where \(r = d \pm in\)

If \(b^2 - 4ac = 0\), \(r_1 = r_2\), so need 2nd (independent) solution:  \(te^{r_1t}\)

Hence general solution is \(y = c_1e^{r_1t} + c_2te^{r_1t}\).

Initial value problem:  use \(y(t_0) = y_0\), \(y'(t_0) = y_0'\) to solve for \(c_1, c_2\) to find unique 
solution. 
Derivation of general solutions:

If \(b^2 - 4ac > 0\) we guessed \(e^{rt}\) is a solution and noted that any linear combination of 
solutions is a solution to a homogeneous linear differential equation.

3.3:  Linear Independence and the Wronskian

Defn: \(f\) and \(g\) are linearly dependent if there exists constants \(c_1, c_2\) 
such that \(c_1 \not= 0\) or \(c_2 \not= 0\) and 

\(c_1f(t) + c_2g(t) = 0\) for all \(t \in (a, b)\)

Thm 3.3.1:  If \(f: (a, b) \rightarrow R\) and \(g(a, b) \rightarrow R\) are differentiable 
functions on (a, b) and if \(W(f, g)(t_0) \not= 0\) for some \(t_0 \in (a, b)\), then \(f\) and \(g\) 
are linearly independent on \((a, b)\).  Moreover, if \(f\) and \(g\) are linearly dependent on $(a, 
b)\(, then \)W(f, g)(t) = 0\(  for all \)t \in (a, b)$

If \(c_1f(t) + c_2g(t) = 0\) for all \(t\),
then
\(c_1f'(t) + c_2g'(t) = 0\)

Solve the following linear system of equations for \(c_1, c_2\)

$\matrix{c_1 f(t_0) + c_2g(t_0) = 0
\cr
c_1 f'(t_0) + c_2g'(t_0) = 0}$

$\left[\matrix{f(t_0) & g(t_0) \cr
f'(t_0) & g'(t_0)}\right]
\left[\matrix{
c_1 \cr c_2
}\right]
=
\left[\matrix{
0 \cr 0
}\right]$

\end

START 100/ch3_134.tex part 2 AND 100/ch3new.tex part 2

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).
 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\).
 
Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

{\bf When does the following IVP have  unique sol'n:}

IVP:  \(ay'' + by' + cy = 0\), \(y(t_0) = y_0\), \(y'(t_0) = y_1\).

Suppose \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is a solution to 
 
\(y' = c_1\phi_1'(t) + c_2\phi_2'(t)\)}

\(y(t_0) = y_0\):  \(y_0 = c_1\phi_1(t_0) + c_2\phi_2(t_0)\)

\(y'(t_0) = y_1\):  \(y_1 = c_1\phi_1'(t_0) + c_2\phi_2'(t_0)\)

To find IVP solution, need to solve above system of two equations for the unknowns \(c_1\) and \(c_2\).

Note the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\).

Note that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0,  \phi_1(t_0), \phi_2(t_0)\), \(y_1, \phi_1'(t_0), \phi_2'(t_0)\) are the constants.  We can translate this linear system of equations into matrix form:
\u
$\matrix{c_1\phi_1(t_0) + c_2\phi_2(t_0) = y_0 \cr
c_1\phi_1'(t_0) + c_2\phi_2'(t_0) = y_1}\( \hfill \)\Rightarrow$ \hfill
$\left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right]\left[\matrix{c_1 \cr c_2} \right]=\left[\matrix{y_0 \cr y_1} \right] $

Note this equation has a unique solution if and only if 
 
$det \left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0) \right] = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|  = \phi_1 \phi_2' - \phi_1'\phi_2 \not= 0$ }

Definition:  The Wronskian of two differential functions, \(\phi_1\) and \(\phi_2\) is 
\u\u
\(W(\phi_1, \phi_2) = \phi_1 \phi_2' - \phi_1'\phi_2 = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'\right|\)}

Examples:
\u\u\u
1.)  W(\(cos(t)\), \(sin(t)\)) =
\(\left|\matrix{cos(t) & sin(t) \cr -sin(t) & cos(t)}\right| \)
  
\(~~~~~~~~~~~~~~~~= cos^2(t) + sin^2(t) = 1 > 0.\) 

2.)  W(\(e^{dt}cos(nt)\), \(e^{dt}sin(nt)) =\) 
  
~~~~$ \left|\matrix{e^{dtcos(nt) && e^{dt}sin(nt) \cr
 de^{dt}cos(nt)- ne^{dt}sin(nt) &~& de^{dt}sin(nt) + ne^{dt}cos(nt)}\right|$}

\(^{= e^{dt}cos(nt)(de^{dt}sin(nt) + ne^{dt}cos(nt) ) - e^{dt}sin(nt)( de^{dt}cos(nt)- ne^{dt}sin(nt) )}\)

\u\u
$~^{= e^{2dt}%\huge
[cos(nt)(dsin(nt) + ncos(nt)) - sin(nt)(dcos(nt)- nsin(nt) )]%\huge
}$
\u\u
$~^{= e^{2dt}%\huge
[dcos(nt)sin(nt) + ncos^2(nt) - dsin(nt)cos(nt)+ nsin^2(nt)]%\huge
)}$
\u\u

$~~= e^{2dt}%\huge
[ ncos^2(nt) + nsin^2(nt)%\huge
]$
  
[ cos^2(nt) + sin^2(nt)%\huge
]$
\(~=~ ne^{2dt} > 0\) for all \(t\).}

Definition:  The Wronskian of two differential functions, \(f\) and \(g\) is 

\(W(f, g) = fg' - f'g = \big|\matrix{f & g \cr f' & g'\big|\)}

Thm 3.2.3:  Suppose that \hfil \break
 \(\phi_1\) and \(\phi_2\) are two solutions to $y'' + p(t) y' + q(t)y = 
0$. \hb
  If \(W(\phi_1, \phi_2)(t_0) = \phi_1(t_0)\phi_2'(t_0) - \phi_1'(t_0)\phi_2(t_0) \not= 0\), then \hb
 there is a unique choice of constants \(c_1\) and \(c_2\) such that \(c_1\phi_1 + c_2\phi_2\) 
satisfies this 
homogeneous linear differential 
equation and initial conditions, \(y(t_0) = y_0\), \(y'(t_0) = y_0'\).

Thm 3.2.4:  Given the hypothesis of thm 3.2.1, \hb
suppose that \(\phi_1\) and \(\phi_2\) are two solutions to 
 
\(y'' + p(t) y' + q(t)y =0\).
 \u\u
 If \(W(\phi_1, \phi_2)(t_0) \not= 0\), 
for some \(t_0 \in (a, b)\), then any solution to this homogeneous linear differential
equation can be written as $y = 
c_1\phi_1 + c_2\phi_2\( for some \)c_1\( and \)c_2$.

Defn  If \(\phi_1\) and \(\phi_2\) satisfy the conditions in thm 3.2.4, then \(\phi_1\) and \(\phi_2\) 
form a fundamental set of solutions to $y'' + p(t) y' + q(t)y =
0$. 

Thm 3.2.5:  Given any second order homogeneous linear differential equation, there exist a pair 
of functions which form a fundamental set of solutions.

3.3:  Linear Independence and the Wronskian

Defn: \(f\) and \(g\) are linearly dependent if there exists constants \(c_1, c_2\) 
such that \(c_1 \not= 0\) or \(c_2 \not= 0\) and 

\(c_1f(t) + c_2g(t) = 0\) for all \(t \in (a, b)\)

Thm 3.3.1:  If \(f: (a, b) \rightarrow R\) and \(g(a, b) \rightarrow R\) are differentiable 
functions on (a, b) and if \(W(f, g)(t_0) \not= 0\) for some \(t_0 \in (a, b)\), then \(f\) and \(g\) 
are linearly independent on \((a, b)\).  Moreover, if \(f\) and \(g\) are linearly dependent on $(a, 
b)\(, then \)W(f, g)(t) = 0\(  for all \)t \in (a, b)$

If \(c_1f(t) + c_2g(t) = 0\) for all \(t\),
then
\(c_1f'(t) + c_2g'(t) = 0\)

Solve the following linear system of equations for \(c_1, c_2\)

$\matrix{c_1 f(t_0) + c_2g(t_0) = 0
\cr
c_1 f'(t_0) + c_2g'(t_0) = 0}$

$\left[\matrix{f(t_0) & g(t_0) \cr
f'(t_0) & g'(t_0)}\right]
\left[\matrix{
c_1 \cr c_2
}\right]
=
\left[\matrix{
0 \cr 0
}\right]$

START 100/3_2.tex

 
{\bf Existence and Uniqueness for  LINEAR DEs.}
\u\u
\underbar{Homogeneous:}
 

\(y^{(n) + p_1(t)y^{(n-1)} + ... p_{n-1}(t) y' + p_n(t)y = 0\)}
\u
\underbar{Non-homogeneous:} \(g(t) \not= 0\)
 

\(y^{(n) + p_1(t)y^{(n-1)} + ... p_{n-1}(t) y' + p_n(t)y = g(t)\)}
\u
{\bf 1st order LINEAR differential equation:}
\u\u

Thm 2.4.1:  If \(p:(a, b) \rightarrow R\) and \(g:(a, b) \rightarrow R\) are continuous and $a < 
t_0 < b$, then 
there exists a unique function \(y = \phi(t)\), ~\(\phi:(a, b) \rightarrow R\) that satisfies the 
 
IVP:  \(y' + p(t) y = g(t)\), ~~\(y(t_0) = y_0\)

\u
Thm:  If \(y = \phi_1(t)\) is a solution to \underbar{homogeneous} equation, \(y' + p(t) y = 0\), then \(y = c\phi_1(t)\) is the general solution to this equation.
\u\u
If in addition \(y = \psi(t)\) is a solution to \underbar{non-homogeneous} equation, \(y' + p(t) y = g(t)\), then \(y = c\phi_1(t) + \psi(t)\) is the general solution to this equation.

Partial proof:  \(y = \phi_1(t)\) is a solution to  \(y' + p(t) y = 0\)
implies

Thus \(y = c\phi_1(t)\) is a solution to  \(y' + p(t) y = 0\) since
 fill

 \(y = \psi(t)\) is a solution to  \(y' + p(t) y = g(t)\)
implies
 fill
Thus \(y = c\phi_1(t) + \psi(t)\) is a solution to  \(y' + p(t) y = g(t)\) since

{\bf 2nd order LINEAR differential equation:}
\u

Thm 3.2.1:  If \(p:(a, b) \rightarrow R\), \(q:(a, b) \rightarrow R\), and \(g:(a, b) \rightarrow R\) 
are 
continuous and \(a < t_0 < b\), then
there exists a unique function \(y = \phi(t)\), \(\phi:(a, b) \rightarrow R\) that satisfies the
initial value problem

\(y'' + p(t) y' + q(t)y = g(t)\),

\(y(t_0) = y_0\),

\(y'(t_0) = y_0'\)

Thm 3.2.2:  If \(\phi_1\) and \(\phi_2\) are two solutions to a \underbar{homogeneous} linear differential 
equation, then 
\(c_1\phi_1 + c_2\phi_2\) is also a solution to this linear differential equation.

Proof of thm 3.2.2:  

Since \(y(t) = \phi_i(t)\) is a solution to the linear homogeneous differential equation \(y'' + py' + qy = 0\) where \(p\) and \(q\) are functions of \(t\) (note this includes the case with constant coefficients), then

Claim:   \(y(t) = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution to \(y'' + py' + qy = 0\) 

Pf of claim:  

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).
 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\).
 
Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

{\bf When does the following IVP have  unique sol'n:}

IVP:  \(ay'' + by' + cy = 0\), \(y(t_0) = y_0\), \(y'(t_0) = y_1\).

Suppose \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is a solution to 
 
\(y' = c_1\phi_1'(t) + c_2\phi_2'(t)\)}

\(y(t_0) = y_0\):  ~~\(y_0 = c_1\phi_1(t_0) + c_2\phi_2(t_0)\)

\(y'(t_0) = y_1\):  ~~\(y_1 = c_1\phi_1'(t_0) + c_2\phi_2'(t_0)\)

To find IVP solution, need to solve above system of two equations for the unknowns \(c_1\) and \(c_2\).

Note the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\).

Note that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0,  \phi_1(t_0), \phi_2(t_0)\), \(y_1, \phi_1'(t_0), \phi_2'(t_0)\) are the constants.  We can translate this linear system of equations into matrix form:
\u
$\matrix{c_1\phi_1(t_0) + c_2\phi_2(t_0) = y_0 \cr
c_1\phi_1'(t_0) + c_2\phi_2'(t_0) = y_1}\( \hfill \)\Rightarrow$ \hfill
$\left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right]\left[\matrix{c_1 \cr c_2} \right]=\left[\matrix{y_0 \cr y_1} \right] $

Note this equation has a unique solution if and only if 
 
$det \left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0) \right] = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|  = \phi_1 \phi_2' - \phi_1'\phi_2 \not= 0$ }

Definition:  The Wronskian of two differential functions, \(\phi_1\) and \(\phi_2\) is 
\u\u
\(W(\phi_1, \phi_2) = \phi_1 \phi_2' - \phi_1'\phi_2 = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'\right|\)}

Examples:
\u\u\u
1.)  W(\(cos(t)\), \(sin(t)\)) =
\(\left|\matrix{cos(t) & sin(t) \cr -sin(t) & cos(t)}\right| \)
  
\(~~~~~~~~~~~~~~~~= cos^2(t) + sin^2(t) = 1 > 0.\) 

2.)  W(\(e^{dt}cos(nt)\), \(e^{dt}sin(nt)) =\) 
  
~~~~$ \left|\matrix{e^{dtcos(nt) && e^{dt}sin(nt) \cr
 de^{dt}cos(nt)- ne^{dt}sin(nt) &~& de^{dt}sin(nt) + ne^{dt}cos(nt)}\right|$}

\(^{= e^{dt}cos(nt)(de^{dt}sin(nt) + ne^{dt}cos(nt) ) - e^{dt}sin(nt)( de^{dt}cos(nt)- ne^{dt}sin(nt) )}\)

\u\u
$~^{= e^{2dt}%\huge
[cos(nt)(dsin(nt) + ncos(nt)) - sin(nt)(dcos(nt)- nsin(nt) )]%\huge
}$
\u\u
$~^{= e^{2dt}%\huge
[dcos(nt)sin(nt) + ncos^2(nt) - dsin(nt)cos(nt)+ nsin^2(nt)]%\huge
)}$
\u\u

$~~= e^{2dt}%\huge
[ ncos^2(nt) + nsin^2(nt)%\huge
]$
  
[ cos^2(nt) + sin^2(nt)%\huge
]$
\(~=~ ne^{2dt} > 0\) for all \(t\).}

Definition:  The Wronskian of two differential functions, \(f\) and \(g\) is 

\(W(f, g) = fg' - f'g = \big|\matrix{f & g \cr f' & g'\big|\)}

Thm 3.2.3:  Suppose that 
 \(\phi_1\) and \(\phi_2\) are two solutions to $y'' + p(t) y' + q(t)y = 
0$. \hb
 There is a unique choice of constants \(c_1\) and \(c_2\) such that \(c_1\phi_1 + c_2\phi_2\) 
satisfies this 
homog linear differential 
equation and initial conditions, \(y(t_0) = y_0\), \(y'(t_0) = y_0'\).

iff
  \(W(\phi_1, \phi_2)(t_0) = \phi_1(t_0)\phi_2'(t_0) - \phi_1'(t_0)\phi_2(t_0) \not= 0\).

 fill
Thm 3.2.4:  Given the hypothesis of thm 3.2.1, \hb
suppose that \(\phi_1\) and \(\phi_2\) are two solutions to 
 
\(y'' + p(t) y' + q(t)y =0\).
 \u\u
 If \(W(\phi_1, \phi_2)(t_0) \not= 0\), 
for some \(t_0 \in (a, b)\), then any solution to this homogeneous linear differential
equation can be written as $y = 
c_1\phi_1 + c_2\phi_2\( for some \)c_1\( and \)c_2$.
 fill
Defn  If \(\phi_1\) and \(\phi_2\) satisfy the conditions in thm 3.2.4, then \(\phi_1\) and \(\phi_2\) 
form a fundamental set of solutions to $y'' + p(t) y' + q(t)y =
0$. 

 fill
Thm 3.2.5:  Given any second order homogeneous linear differential equation, there exist a pair 
of functions which form a fundamental set of solutions.

FYI:  Linear Independence and the Wronskian

Defn: \(\po\) and \(\pt\) are linearly dependent if there exists constants \(c_1, c_2\) 
such that \(c_1 \not= 0\) or \(c_2 \not= 0\) and 

\(c_1\po(t) + c_2\pt(t) = 0\) for all \(t \in (a, b)\)

Thm 3.3.1:  If \(\po: (a, b) \rightarrow R\) and \(\pt(a, b) \rightarrow R\) are differentiable 
functions on (a, b) and \hb if \(W(\po, \pt)(t_0) \not= 0\) for some \(t_0 \in (a, b)\), then\hb \(\po\) and \(\pt\) 
are linearly independent on \((a, b)\).  \hb Moreover, if \(\po\) and \(\pt\) are linearly dependent on $(a, 
b)\(, then \)W(\po, \pt)(t) = 0\(  for all \)t \in (a, b)$

Proof idea:

If \(c_1\po(t) + c_2\pt(t) = 0\) for all \(t\in (a, b)\),

\(c_1\po'(t) + c_2\pt'(t) = 0\) for all \(t\in (a, b)\)}

Solve the following linear system of equations for \(c_1, c_2\)

$\matrix{c_1 \po(t_0) + c_2\pt(t_0) = 0
\cr
c_1 \po'(t_0) + c_2\pt'(t_0) = 0}$

$\left[\matrix{\po(t_0) & \pt(t_0) \cr
\po'(t_0) & \pt'(t_0)}\right]
\left[\matrix{
c_1 \cr c_2
}\right]
=
\left[\matrix{
0 \cr 0
}\right]$
 fill
In other words the fundamental set of solutions \(\{\po, \pt\}\) to $y'' + p(t) y' + q(t)y =
0$ form a basis for the set of all solutions to this linear homogeneous DE.

\end

Thm:  Suppose \(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

If \(\psi\) is a solution to

\(ay'' + by' + cy = g(t)\) [*],
Then \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is also a solution to [*].

Moreover if \(\gamma\) is also a solution to [*], then there exist constants \(c_1, c_2\) such that 

\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\)

Or in other words, \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to [*].

Proof:  

Define   \(L(f) = af'' + bf' + cf\).

Recall \(L\) is a linear function.

 Let \(h = c_1 \phi_1(t) + c_2 \phi_2(t)\).  Since \(h\) is a solution to the differential 
equation, \(ay'' + by' + cy = 0\),

Since \(\psi\) is a solution to 
 \(ay'' + by' + cy = g(t)\),

We will now show that \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h\) is also a solution to 
[*].

Since \(\gamma\) a solution to  \(ay'' + by' + cy = g(t)\),

We will first show that \(\gamma - \psi\) is a solution to the differential equation
 \(ay'' + by' + cy = 0\).

 fill

Since \(\gamma - \psi\) is a solution to \(ay'' + by' + cy = 0\) and 

\(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

there exist constants \(c_1, c_2\) such that

\(\gamma - \psi =\underline{\hskip 2in\)}

Thus
\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\).

Thm: \hfil \break
 Suppose  
\(f_1\) is a a solution to
\(ay'' + by' + cy = g_1(t)\)  \hfil \break
and \(f_2\) is a a solution to
\(ay'' + by' + cy = g_2(t)\), then \(f_1 + f_2\) is a solution to
\(ay'' + by' + cy = g_1(t) + g_2(t)\)

Proof: Let  \(L(f) = af'' + bf' + cf\).

Since \(f_1\) is a solution to 
 \(ay'' + by' + cy = g_1(t)\),

Since \(f_2\) is a solution to 
 \(ay'' + by' + cy = g_2(t)\),

We will now show that \(f_1 + f_2\) is a solution to \break
 \(ay'' + by' + cy = g_1(t) + g_2(t)\).

 fill

Sidenote:  The proofs above work even if \(a, b, c\) are functions of \(t\) instead of constants.

{\bf Examples:} Find a suitable form for \(\psi\) for the following differential equations:

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 fill

2.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 fill

3.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 fill

4.)  \(y'' - 5y  = 4sin(3t)\)
 fill

5.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 fill

6.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 fill

7.)  
\(y'' - 4y' - 5y = 4 sin(3t) e^{2t}\)
 fill
8.)  
\(y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}\)
 fill

9.) \(y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}\)
 fill
10.)  \(y'' - 4y' - 5y\)

 fill
11.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 fill
12.) \(y'' - 4y' - 5y = 4e^{-t}\)
 fill
To solve \(ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)\) [**]

1.)  Find the general solution to \(ay'' + by' + cy = 0\):  

\(c_1\phi_1 + c_2\phi_2\)

2.)  For each \(g_i\), find a solution to \(ay'' + by' + cy = g_i\):  

\(\psi_i\)

This includes plugging guessed solution {\(\psi_i\)} into \break
\(ay'' + by' + cy = g_i\).

The general solution to [**] is

\(c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n\)

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.) {\bf First solve homogeneous equation:}

  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to \hb
  LINEAR HOMOGENEOUS equation.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.) {\bf Find one solution to non-homogeneous eq'n:}
  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)
~~~~~~~~\(y' = 3A cos(3t) - 3B sin(3t)\)
~~~~~~~~\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.) {\bf If initial value problem:}
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

\((e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0\) and 
\((e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0\) 

Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/3_2a.tex

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 etlength{\textwidth}{7.4in}
 etlength{\textheight}{9.6in}
\pagestyle{empty}  %% To avoid page numbering
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{relsize}

\AppendGraphicsExtensions{.gif}
\DeclareGraphicsExtensions{.pdf,.png,.gif,.jpg}
\begin{document}

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).
 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\).
Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

{\bf When does the following IVP have a unique solution:}

IVP:  \(ay'' + by' + cy = 0\), \(y(t_0) = y_0\), \(y'(t_0) = y_1\).

Suppose \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is a solution to \(ay'' + by' + cy = 0\).  Then 
\(y' = c_1\phi_1'(t) + c_2\phi_2'(t)\)

\(y(t_0) = y_0\):  \(y_0 = c_1\phi_1(t_0) + c_2\phi_2(t_0)\)

\(y'(t_0) = y_1\):  \(y_1 = c_1\phi_1'(t_0) + c_2\phi_2'(t_0)\)

To find IVP solution, need to solve above system of two equations for the unknowns \(c_1\) and \(c_2\).

Note the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\).

Note that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0,  \phi_1(t_0), \phi_2(t_0)\), \(y_1, \phi_1'(t_0), \phi_2'(t_0)\) are the constants.  We can translate this linear system of equations into matrix form:

$\matrix{c_1\phi_1(t_0) + c_2\phi_2(t_0) = y_0 \cr
c_1\phi_1'(t_0) + c_2\phi_2'(t_0) = y_1}$ ~~~implies~~~
$\left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right]
\left[\matrix{c_1 \cr c_2} \right] = \left[\matrix{y_0 \cr y_1} \right] $

Note this equation has a unique solution if and only if $det \left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right] = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|  = \phi_1 \phi_2' - \phi_1'\phi_2 \not= 0$ 

Definition:  The Wronskian of two differential functions, \(\phi_1\) and \(\phi_2\) is 

\(W(\phi_1, \phi_2) = \phi_1 \phi_2' - \phi_1'\phi_2 = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'\right|\)}

Examples:

1.)  Wronskian of \(cos(t)\), \(sin(t)\) =
\(\left|\matrix{cos(t) & sin(t) \cr -sin(t) & cos(t)}\right| = cos^2(t) + sin^2(t) = 1 > 0.\) 

2.)  Wronskian of \(e^{dt}cos(nt)\), \(e^{dt}sin(nt)\) =
$\left|\matrix{e^{dt}cos(nt) && e^{dt}sin(nt) \cr
 de^{dt}cos(nt)- ne^{dt}sin(nt) &~& de^{dt}sin(nt) + ne^{dt}cos(nt)}\right|$}

$= e^{dt}cos(nt)[de^{dt}sin(nt) + ne^{dt}cos(nt)] - 
e^{dt}sin(nt)[ de^{dt}cos(nt)- ne^{dt}sin(nt)]$

\(= e^{2dt}\huge(cos(nt)[dsin(nt) + ncos(nt)] - sin(nt)[ dcos(nt)- nsin(nt)]\huge)\)

\(= e^{2dt}\huge(dcos(nt)sin(nt) + ncos^2(nt)] - dsin(nt)cos(nt)+ nsin^2(nt)]\huge)\)

\(= e^{2dt}\huge( ncos^2(nt) + nsin^2(nt)\huge)\)
\(= ne^{2dt}\huge( cos^2(nt) + sin^2(nt)\huge)\)
\(= ne^{2dt} > 0\) for all \(t\).
\end{document}

START 100/FALL16/e1_Fall2016ANS.tex part 6

[20]~ 1.)  Solve \(y'' - 6y' + 9 = 0\), \(y(0) = 2\), \(y'(0) = 4\).

 fill
Answer: \(\underline{~~\hskip 4in~~\)}

\end

START 100/FALL18/final3600_F2016ans.tex part 3

\documentclass[12pt]{article}

 etlength{\topmargin}{-1in}
 etlength{\oddsidemargin}{-.250in}
 etlength{\textwidth}{7.0in}
 etlength{\textheight}{10.1in}
\pagestyle{empty}  %% To avoid page numbering
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{relsize}

\begin{document}

MATH:3600:0002 Final Exam \hfil \break
Dec. 12, 2016 \hfill SHOW ALL STEPS \hfill

[6] ~1.)  The  Wronskian of \(e^{-t}cos(t)\),  \(e^{-t}sin(t)\) is 
~\(W(e^{-t}cos(t),  e^{-t}sin(t)) = \underline{~~e^{-2t}~~}\)

$W(e^{-t}cos(t),  e^{-t}sin(t)) = \left|\matrix{e^{-t}cos(t) &  e^{-t}sin(t) \cr 
 - e^{-t}sin(t)  - e^{-t}cos(t) &  e^{-t}cos(t)  -e^{-t}sin(t)}\right|$
 
\hskip 1.3in $= e^{-t}cos(t)[e^{-t}cos(t)  -e^{-t}sin(t)]
- e^{-t}sin(t) [ -e^{-t}sin(t)  - e^{-t}cos(t) ]$
 
\hskip 1.3in $= e^{-2t}cos^2(t)  -e^{-2t}cos(t)sin(t)
+  e^{-2t}sin^2(t)  + e^{-2t}cos(t) sin(t)$
 
\hskip 1.3in $= e^{-2t}[cos^2(t) 
+  sin^2(t)] = e^{-2t}$

\end

START 34/FALL03/finalreviewANS2.txt part 2

Note the following review problems DO NOT cover all problem types which may appear on the 
final.  

Chapter 3:

9.)  Solve the following initial problems:

9a.)  \(y'' + 6y' + 8y = 0, ~y(0) = 0, ~y'(0) = 0 \)

Suppose \(y = e^{rt}\).  Then \(y' = re^{rt}, ~y'' = r^2e^{rt}\)

\(r^2e^{rt} + 6re^{rt} + 8e^{rt} = 0\)
Hence \(r^2 + 6r + 8 = 0\).  Thus, (r + 2)(r + 4) = 0\(.  Hence \)r = -2, -4$

Hence general solution is \(y(t) = c_1e^{-2t} + c_2e^{-4t}\)

\(y(0) = 0:  0 = c_1 + c_2\).  Thus \(c_2 = -c_1\)

\(y'(0) = 0:  y' = -2c_1e^{-2t} - 4c_2e^{-4t}\)

\(0 = -2c_1 - 4c_2 = 2c_2 - 4c_2 = -2c_2\)  Thus \(c_2 = 0, c_1 = 0\)

Thus, \(y(t) = 0\) is the solution to the initial value problem.

9b.)  \(y'' + 6y' + 9y = 0, ~y(0) = 0, ~y'(0) = 0 \)

\(r^2 + 6r + 9 = (r + 3)(r +3) = 0\)

Hence general solution is \(y(t) = c_1e^{-3t} + c_2te^{-3t}\)

\(y(0) = 0:  0 = c_1 + c_2(0)\).  Thus \(c_1 = 0\)

\(y = c_2te^{-3t}\)

\(y' = c_2[e^{-3t} -3t e^{-3t}]\)

\(y'(0) = 0:  0 =  c_2[e^{0} -3(0) e^{0}] \).  Thus \(c_2 = 0\).

Thus, \(y(t) = 0\) is the solution to the initial value problem.

9c.)  \(y'' + 6y' + 10y = 0, ~y(0) = 0, ~y'(0) = 0\)

\(r^2 + 6r + 10 = 0\), $r = {-6 \pm  qrt{36 - 4(1)(10)} \over 2}  
= {-6 \pm 2i \over 2} = -3 \pm i$

Hence general solution is  \(y(t) = c_1e^{-3t}sin(t) + c_2e^{-3t}cos(t)\)

\(y(0) = 0:  0 = c_1(0) + c_2(1)\).  Thus \(c_2 = 0\)

\(y = c_1 e^{-3t}sin(t) \)

$y' = c_1[-3 e^{-3t}sin(t) +  e^{-3t}cos(t)] 

\(y'(0) = 0:  0 =  c_1[0 + 1] \).  Thus \(c_1 = 0\).

Thus, \(y(t) = 0\) is the solution to the initial value problem.

Quicker method to solve 9a, b, c:

Note that \(y = 0\) was the obvious solution to the initial value problems in 9a, b, c.

Clearly the constant function \(y(t) = 0\) satisfies \(ay'' + by' + cy = 0\) for any 
\(a,~ b,~ c\).  It also satisfies \(y(0) = 0, ~y'(0) = 0\).  Thus this constant function 
is a solution to IVP of this type.  Since  IVP of this type have unique solutions, $y(t) 
= 0$ is the only solution.

\end

START quiz5Fall2017ans.tex part 1

\nopagenumbers

Quiz 5 Form A \hfil \break
Oct 30, 2017

[10]~~ 1.  Let \(y_1\) and \(y_2\) be solutions of
$ty''
+ 4y'
+
sin(t)
y = 0\(; \)t > 0$.
Let \(W(t)\) be the Wronskian of \(y_1(t)\) and \(y_2(t)\). Given that \(W(1) = 3\), find
\(W(t)\).

{\(W(t) = \underline{~~3t^{-4}~~}\)}

\(p(t) = {4 \over t}\).  Thus by Abel's thm \(W(t) =ce^{-\int  {4 dt \over t}}= ce^{-4\int  {dt \over t}}= ce^{-4 ln|t|} =e^{ln|t|^{-4}} = ct^{-4}\) .

Thus  \(W(t) =ct^{-4} \) for some constant \(c\).  Since \(W(1) = 3\),   ~~\(3 =  c(1)^{-4}\).  Thus \(c = 3\).

Hence \(W(t) =3t^{-4} \)

--- 

\end

START 100/FALL16/e2_3600_F2016ANS part 1

[20]~ 4.)  Solve ~~ \(x^2y'' + 8xy' -8y = 0\).

Guess \(y = x^r\).  Then \(y' = rx^{r-1}\) and \(y'' = r(r-1)x^{r-2}\)

Plug \(y = x^r\) into \(x^2y'' + 8xy' -8y = 0\).

\(x^2 r(r-1)x^{r-2} + 8x rx^{r-1} -8x^r = 0\).

\((r^2 -r)x^{r} + 8rx^{r} -8x^r = 0\).

\((r^2 -r + 8r -8)x^r = 0\).

\((r^2 + 7r -8)x^r = 0\) implies \((r-1)(r + 8) = 0\). Thus \(r = 1, -8\)

Thus \(y = x\) and \(y = x^{-8}\) are two linearly independent solutions to this 2nd order linear homogeneous DE.
Answer: \(\underline{~~y = c_1x + c_2x^{-8~~}\)}

START 100/3_3.tex

\documentclass[12pt]{article}

 etlength{\topmargin}{-0.5in}
 etlength{\oddsidemargin}{-.250in}
 etlength{\textwidth}{7.4in}
 etlength{\textheight}{9.4in}
\pagestyle{empty}  %% To avoid page numbering
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{relsize}

\AppendGraphicsExtensions{.gif}
\DeclareGraphicsExtensions{.pdf,.png,.gif,.jpg}
\begin{document}

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).

RECOMMENDED Method: 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\)

Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

NOT RECOMMENDED:  work with  \(y = c_1e^{it} + c_2 e^{-it}\)

 \(y' = ic_1e^{it} - ic_2 e^{-it}\)

 \(y(0) = -1\):   \(-1 = c_1e^{0} + c_2 e^{0}\) implies \(-1 = c_1 + c_2\).

 \(y'(0) = -3\):   \(-3 = ic_1e^{0} - ic_2 e^{0}\) implies $-3 = ic_1 - 
ic_2$.

 \(-1i = ic_1 + ic_2\).

 \(-3 = ic_1 - ic_2\).

\(2ic_1 = -3 - i\) implies \(c_1 = {-3i - i^2 \over -2} =  {3i - 1\over 2}\)

\(2ic_2 = 3 - i\) implies \(c_2 = {3i - i^2 \over -2} =  {-3i - 1\over 2}\)

 Euler's formula:  \(e^{ix} = cos(x) + isin(x)\)
 
 $y = ({3i - 1\over 2})e^{it} + ({-3i - 1\over 2}) e^{-it} =
 ({3i - 1\over 2})[cos(t) + isin(t)] + ({-3i - 1\over 2}) [cos(-t) + 
isin(-t)]$
 
 
 $= ({3i - 1\over 2})[cos(t) + isin(t)] + ({-3i - 1\over 2}) [cos(t) - 
isin(t)]$
 
 
  
 $= ({3i \over 2})cos(t) + ({3i \over 2})isin(t) 
+ ({ -1\over 2})cos(t) + ({-1 \over 2})isin(t)
 + ({-3i \over 2}) cos(t) - ({-3i \over 2})isin(t)
  + ({-1\over 2}) cos(t) - ({-1 \over 2})isin(t)
 $
 
 $=  ({3i \over 2})isin(t) 
+ ({ -1\over 2})cos(t) +  ({3i \over 2})isin(t)
  + ({-1\over 2}) cos(t) 
 $
 
$=  -({3 \over 2})sin(t) 
- ({ 1\over 2})cos(t) -  ({3 \over 2})sin(t)
  - ({1\over 2}) cos(t) 
 $
 
 $=  -{3 }sin(t) 
- { 1}cos(t)
 $
 
 
 
\end{document}

START 100/ch3.tex part 2

Section 3.3:  If \(b^2 - 4ac < 0\), :

Changed format of \(y = c_1e^{r_1t} + c_2e^{r_2t}\) to linear combination of real-valued 
functions 
instead of
complex valued functions by using Euler's formula:

\(e^{it = cos(t) + i sin(t)\)}

Hence \(e^{(d + in)t} =e^{dt}e^{int} = e^{dt}[cos (nt) + i sin (nt)]\)

Let \(r_1 = d + in\), \(r_2 = d - in\)

\(y = c_1e^{r_1t} + c_2e^{r_2t} \)
$= 
c_1e^{dt}[cos (nt) + i sin (nt)] +c_2e^{dt}[cos (-nt) + i sin (-nt)] $
\(=c_1e^{dt}cos (nt) + ic_1 e^{dt} sin (nt) + c_2e^{dt}cos (nt) - ic_2 e^{dt} sin (nt) \)

=\((c_1 + c_2)e^{dt}cos (nt) + i(c_1 - c_2) e^{dt}sin (nt)\)
\hskip 28pt\(=k_1 e^{dt} cos (nt) + k_2e^{dt} sin (nt)\)

\end

START 100/ch3_134.tex part 3 AND 100/ch3new.tex part 3

Ex 2:  Solve \(y'' - 3y' + 4y = 0\).  
\u
\(y = e^{rt}\) implies \(r^2 - 3r + 4= 0\) and hence 

\(r = {3 \pm  qrt{(-3)^2 -4(1)(4)} \over 2}= {3 \over 2}\pm { qrt{9 -16} \over 2} = {3 \over 2}\pm i{ qrt{7} \over 2}\) 

Hence general sol'n is \(y = c_1e^{{3 \over 2t}cos({ qrt{7} \over 2}t)  + c_2e^{{3 \over 2}t}sin({ qrt{7} \over 2}t)\)}

---
\u

Section 3.3:  If \(b^2 - 4ac < 0\), :

Changed format of \(y = c_1e^{r_1t} + c_2e^{r_2t}\) to linear combination of real-valued 
functions 
instead of
complex valued functions by using Euler's formula:

\(e^{it = cos(t) + i sin(t)\)}

Hence \(e^{(d + in)t} =e^{dt}e^{int} = e^{dt}[cos (nt) + i sin (nt)]\)

Let \(r_1 = d + in\), \(r_2 = d - in\)

\(y = c_1e^{r_1t} + c_2e^{r_2t} \)
$= 
c_1e^{dt}[cos (nt) + i sin (nt)] +c_2e^{dt}[cos (-nt) + i sin (-nt)] $
\(=c_1e^{dt}cos (nt) + ic_1 e^{dt} sin (nt) + c_2e^{dt}cos (nt) - ic_2 e^{dt} sin (nt) \)

=\((c_1 + c_2)e^{dt}cos (nt) + i(c_1 - c_2) e^{dt}sin (nt)\)
\hskip 28pt\(=k_1 e^{dt} cos (nt) + k_2e^{dt} sin (nt)\)

\end

START 100/ch7and9.tex part 1

{\bf Section 3.3:}  If \(b^2 - 4ac < 0\), :

Changed format of \(y = c_1e^{r_1t} + c_2e^{r_2t}\) to linear combination of real-valued
functions
instead of
complex valued functions by using Euler's formula:

\(e^{it = cos(t) + i sin(t)\)}

Hence \(e^{(d + in)t} =e^{dt}e^{int} = e^{dt}[cos (nt) + i sin (nt)]\)

Let \(r_1 = d + in\), \(r_2 = d - in\)

\(y = c_1e^{r_1t} + c_2e^{r_2t} = c_1e^{( d + in)t} + c_2e^{( d - in)t} = c_1e^{dt}e^{int} + c_2e^{dt}e^{- int} \)
$=
c_1e^{dt}[cos (nt) + i sin (nt)] +c_2e^{dt}[cos (-nt) + i sin (-nt)] $
\(=c_1e^{dt}cos (nt) + ic_1 e^{dt} sin (nt) + c_2e^{dt}cos (nt) - ic_2 e^{dt} sin (nt) \)

=\((c_1 + c_2)e^{dt}cos (nt) + i(c_1 - c_2) e^{dt}sin (nt)\)
\hskip 28pt$=k_1 e^{dt} cos (nt) + k_2e^{dt} sin (nt)
= e^{dt} [k_1cos (nt) + k_2sin (nt)]$

\end

START 100/3_3.tex

\documentclass[12pt]{article}

 etlength{\topmargin}{-0.5in}
 etlength{\oddsidemargin}{-.250in}
 etlength{\textwidth}{7.4in}
 etlength{\textheight}{9.4in}
\pagestyle{empty}  %% To avoid page numbering
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{relsize}

\AppendGraphicsExtensions{.gif}
\DeclareGraphicsExtensions{.pdf,.png,.gif,.jpg}
\begin{document}

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).

RECOMMENDED Method: 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\)

Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

NOT RECOMMENDED:  work with  \(y = c_1e^{it} + c_2 e^{-it}\)

 \(y' = ic_1e^{it} - ic_2 e^{-it}\)

 \(y(0) = -1\):   \(-1 = c_1e^{0} + c_2 e^{0}\) implies \(-1 = c_1 + c_2\).

 \(y'(0) = -3\):   \(-3 = ic_1e^{0} - ic_2 e^{0}\) implies $-3 = ic_1 - 
ic_2$.

 \(-1i = ic_1 + ic_2\).

 \(-3 = ic_1 - ic_2\).

\(2ic_1 = -3 - i\) implies \(c_1 = {-3i - i^2 \over -2} =  {3i - 1\over 2}\)

\(2ic_2 = 3 - i\) implies \(c_2 = {3i - i^2 \over -2} =  {-3i - 1\over 2}\)

 Euler's formula:  \(e^{ix} = cos(x) + isin(x)\)
 
 $y = ({3i - 1\over 2})e^{it} + ({-3i - 1\over 2}) e^{-it} =
 ({3i - 1\over 2})[cos(t) + isin(t)] + ({-3i - 1\over 2}) [cos(-t) + 
isin(-t)]$
 
 
 $= ({3i - 1\over 2})[cos(t) + isin(t)] + ({-3i - 1\over 2}) [cos(t) - 
isin(t)]$
 
 
  
 $= ({3i \over 2})cos(t) + ({3i \over 2})isin(t) 
+ ({ -1\over 2})cos(t) + ({-1 \over 2})isin(t)
 + ({-3i \over 2}) cos(t) - ({-3i \over 2})isin(t)
  + ({-1\over 2}) cos(t) - ({-1 \over 2})isin(t)
 $
 
 $=  ({3i \over 2})isin(t) 
+ ({ -1\over 2})cos(t) +  ({3i \over 2})isin(t)
  + ({-1\over 2}) cos(t) 
 $
 
$=  -({3 \over 2})sin(t) 
- ({ 1\over 2})cos(t) -  ({3 \over 2})sin(t)
  - ({1\over 2}) cos(t) 
 $
 
 $=  -{3 }sin(t) 
- { 1}cos(t)
 $
 
 
 
\end{document}

START 100/SPRING13/e1_2013ANS.txt part 3

5.)   Find the general solutions for the following three  differential equations.

[15]~  5A.)  \(2y''  - 3y' + 5y = 0\)

\(y = e^{rt}\).  Then \(y' = re^{rt}\) and \(y'' = r^2e^{rt}\).

\(2r^2e^{rt}  - 3re^{rt} + 5e^{rt} = 0\) implies \(2r^2 - 3r + 5 = 0\)

\(2r^2  - 3r + 5 = 0\) implies $r = {3 \pm  qrt{9 - 4(2)(5)} \over 4} = 
{3 \pm  qrt{9 - 40} \over 4} = {3 \pm  qrt{-31} \over 4}
 = {3 \pm  qrt{-31} \over 4} = {3\over 4} \pm i{ qrt{31} \over 4}$

 fill  fill

Answer 5A.) $\underline{~~
y = c_1 e^{3t \over 4cos({ qrt{31} \over 4}~t) + c_2 e^{3t \over 4}sin({ qrt{31} \over 4}~t)
~~}$}
 fill  fill

\end

START 100/ch3.tex part 3

Section 3.4:  If \(b^2 - 4ac = 0\), then \(r_1 = r_2\).  \hfil \break
Hence one solution is \(y = e^{r_1t}\)  Need second 
solution.
\u
If \(y = e^{rt}\) is a solution, \(y = ce^{rt}\) is a solution.
How about \(y = v(t)e^{rt\)?}
\u\u
\(y' = v'(t)e^{rt} + v(t)re^{rt}\)
\u\u
\(y'' = v''(t)e^{rt} +  v'(t)re^{rt} + v'(t)re^{rt} +  v(t)r^2e^{rt}\)
\u\u
$\hskip 14pt = 
v''(t)e^{rt} +  2v'(t)re^{rt} +  v(t)r^2e^{rt} 
$

\(ay'' + by' + cy = 0\)
\u
{\(a(v''e^{rt} + 2v're^{rt} + v r^2e^{rt}) + b(v'e^{rt} + vre^{rt}) + cve^{rt}= 0\)}
\u
$a(v''(t) + 2v'(t)r + v(t)r^2) + b(v'(t) + v(t)r) + cv(t)
= 0$
\u
$av''(t) + 2av'(t)r + av(t)r^2 + bv'(t) + bv(t)r + cv(t)
= 0$
\u

$av''(t) 
+ (2ar+b)v'(t) 
+ (ar^2
+ br + c)v(t)
= 0$
\u
$av''(t) 
+ (2a({-b \over 2a})+b)v'(t) 
+ 0
= 0$

\(ar^2 + br + c = 0\)
and \(r = {-b \over 2a}\)}
\u

$av''(t) 
+ (-b + b)v'(t) 
= 0$.
\hfill
Thus
\(av''(t)  = 0\).
\u
Hence 
\(v''(t)  = 0\)
and
\(v'(t)  = k_1\)
and
\(v(t)  = k_1t + k_2\)

Hence \(v(t)e^{r_1t   = (k_1t + k_2)e^{r_1t}\) is a soln}
Thus \(te^{r_1t}\) is a nice second solution.
\u
Hence general solution is \(y = c_1 e^{r_1t}+ c_2 te^{r_1t}\)

Solve: \(y'' + y = 0\),  \(y(0) = -1\), \(y'(0) = -3\)

\(r^2 + 1 = 0\) implies \(r^2 = -1\).  Thus \(r = \pm i\).
 

 Since \(r = 0 \pm 1i\),  \(y = k_1cos(t) + k_2sin(t)\).
 
Then \(y' = -k_1sin(t) + k_2 cos(t)\)

 \(y(0) = -1\):      \(-1 = k_1cos(0) + k_2sin(0)\) implies \(-1 = k_1\)

 \(y'(0) = -3\):   \(-3 = -k_1sin(0) + k_2 cos(0)\) implies \(-3 = k_2\)

Thus IVP solution:  \(y = -cos(t)  - 3sin(t)\)

{\bf When does the following IVP have  unique sol'n:}

IVP:  \(ay'' + by' + cy = 0\), \(y(t_0) = y_0\), \(y'(t_0) = y_1\).

Suppose \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is a solution to 
 
\(y' = c_1\phi_1'(t) + c_2\phi_2'(t)\)}

\(y(t_0) = y_0\):  \(y_0 = c_1\phi_1(t_0) + c_2\phi_2(t_0)\)

\(y'(t_0) = y_1\):  \(y_1 = c_1\phi_1'(t_0) + c_2\phi_2'(t_0)\)

To find IVP solution, need to solve above system of two equations for the unknowns \(c_1\) and \(c_2\).

Note the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\).

Note that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0,  \phi_1(t_0), \phi_2(t_0)\), \(y_1, \phi_1'(t_0), \phi_2'(t_0)\) are the constants.  We can translate this linear system of equations into matrix form:
\u
$\matrix{c_1\phi_1(t_0) + c_2\phi_2(t_0) = y_0 \cr
c_1\phi_1'(t_0) + c_2\phi_2'(t_0) = y_1}\( \hfill \)\Rightarrow$ \hfill
$\left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0)} \right]\left[\matrix{c_1 \cr c_2} \right]=\left[\matrix{y_0 \cr y_1} \right] $

Note this equation has a unique solution if and only if 
 
$det \left[\matrix{ \phi_1(t_0) & \phi_2(t_0) \cr
\phi_1'(t_0) &\phi_2'(t_0) \right] = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|  = \phi_1 \phi_2' - \phi_1'\phi_2 \not= 0$ }

Definition:  The Wronskian of two differential functions, \(\phi_1\) and \(\phi_2\) is 
\u\u
\(W(\phi_1, \phi_2) = \phi_1 \phi_2' - \phi_1'\phi_2 = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'\right|\)}

Examples:
\u\u\u
1.)  W(\(cos(t)\), \(sin(t)\)) =
\(\left|\matrix{cos(t) & sin(t) \cr -sin(t) & cos(t)}\right| \)
  
\(~~~~~~~~~~~~~~~~= cos^2(t) + sin^2(t) = 1 > 0.\) 

2.)  W(\(e^{dt}cos(nt)\), \(e^{dt}sin(nt)) =\) 
  
~~~~$ \left|\matrix{e^{dtcos(nt) && e^{dt}sin(nt) \cr
 de^{dt}cos(nt)- ne^{dt}sin(nt) &~& de^{dt}sin(nt) + ne^{dt}cos(nt)}\right|$}

\(^{= e^{dt}cos(nt)(de^{dt}sin(nt) + ne^{dt}cos(nt) ) - e^{dt}sin(nt)( de^{dt}cos(nt)- ne^{dt}sin(nt) )}\)

\u\u
$~^{= e^{2dt}%\huge
[cos(nt)(dsin(nt) + ncos(nt)) - sin(nt)(dcos(nt)- nsin(nt) )]%\huge
}$
\u\u
$~^{= e^{2dt}%\huge
[dcos(nt)sin(nt) + ncos^2(nt) - dsin(nt)cos(nt)+ nsin^2(nt)]%\huge
)}$
\u\u

$~~= e^{2dt}%\huge
[ ncos^2(nt) + nsin^2(nt)%\huge
]$
  
[ cos^2(nt) + sin^2(nt)%\huge
]$
\(~=~ ne^{2dt} > 0\) for all \(t\).}

Definition:  The Wronskian of two differential functions, \(f\) and \(g\) is 

\(W(f, g) = fg' - f'g = \big|\matrix{f & g \cr f' & g'\big|\)}

Thm 3.2.3:  Suppose that \hfil \break
 \(\phi_1\) and \(\phi_2\) are two solutions to $y'' + p(t) y' + q(t)y = 
0$. \hb
  If \(W(\phi_1, \phi_2)(t_0) = \phi_1(t_0)\phi_2'(t_0) - \phi_1'(t_0)\phi_2(t_0) \not= 0\), then \hb
 there is a unique choice of constants \(c_1\) and \(c_2\) such that \(c_1\phi_1 + c_2\phi_2\) 
satisfies this 
homogeneous linear differential 
equation and initial conditions, \(y(t_0) = y_0\), \(y'(t_0) = y_0'\).

Thm 3.2.4:  Given the hypothesis of thm 3.2.1, \hb
suppose that \(\phi_1\) and \(\phi_2\) are two solutions to 
 
\(y'' + p(t) y' + q(t)y =0\).
 \u\u
 If \(W(\phi_1, \phi_2)(t_0) \not= 0\), 
for some \(t_0 \in (a, b)\), then any solution to this homogeneous linear differential
equation can be written as $y = 
c_1\phi_1 + c_2\phi_2\( for some \)c_1\( and \)c_2$.

Defn  If \(\phi_1\) and \(\phi_2\) satisfy the conditions in thm 3.2.4, then \(\phi_1\) and \(\phi_2\) 
form a fundamental set of solutions to $y'' + p(t) y' + q(t)y =
0$. 

Thm 3.2.5:  Given any second order homogeneous linear differential equation, there exist a pair 
of functions which form a fundamental set of solutions.

\end

START 100/ch3_134.tex part 4 AND 100/ch3new.tex part 4

Ex 3:  \(y'' - 6y' + 9y = 0\)  implies \(r^2 - 6r + 9 = (r-3)^2= 0\) 
\u\u\u
Repeated root, \(r = 3\) implies  

general solution is \(y = c_1e^{3t  + c_2te^{3t}\)}

So why did we guess \(y = e^{rt}\)?

Goal:  Solve linear homogeneous 2nd order DE with constant coefficients, 

\(ay'' + by' + cy = 0\) where \(a, b, c\) are constants

Standard mathematical technique:  make up simpler problems and see if you can generalize to the problem of interest.

Section 3.4:  If \(b^2 - 4ac = 0\), then \(r_1 = r_2\).  \hfil \break
Hence one solution is \(y = e^{r_1t}\)  Need second 
solution.
\u
If \(y = e^{rt}\) is a solution, \(y = ce^{rt}\) is a solution.
How about \(y = v(t)e^{rt\)?}
\u\u
\(y' = v'(t)e^{rt} + v(t)re^{rt}\)
\u\u
\(y'' = v''(t)e^{rt} +  v'(t)re^{rt} + v'(t)re^{rt} +  v(t)r^2e^{rt}\)
\u\u
$\hskip 14pt = 
v''(t)e^{rt} +  2v'(t)re^{rt} +  v(t)r^2e^{rt} 
$

\(ay'' + by' + cy = 0\)
\u
{\(a(v''e^{rt} + 2v're^{rt} + v r^2e^{rt}) + b(v'e^{rt} + vre^{rt}) + cve^{rt}= 0\)}
\u
$a(v''(t) + 2v'(t)r + v(t)r^2) + b(v'(t) + v(t)r) + cv(t)
= 0$
\u
$av''(t) + 2av'(t)r + av(t)r^2 + bv'(t) + bv(t)r + cv(t)
= 0$
\u

$av''(t) 
+ (2ar+b)v'(t) 
+ (ar^2
+ br + c)v(t)
= 0$
\u
$av''(t) 
+ (2a({-b \over 2a})+b)v'(t) 
+ 0
= 0$

\(ar^2 + br + c = 0\)
and \(r = {-b \over 2a}\)}
\u

$av''(t) 
+ (-b + b)v'(t) 
= 0$.
\hfill
Thus
\(av''(t)  = 0\).
\u
Hence 
\(v''(t)  = 0\)
and
\(v'(t)  = k_1\)
and
\(v(t)  = k_1t + k_2\)

Hence \(v(t)e^{r_1t   = (k_1t + k_2)e^{r_1t}\) is a soln}
Thus \(te^{r_1t}\) is a nice second solution.
\u
Hence general solution is \(y = c_1 e^{r_1t}+ c_2 te^{r_1t}\)

{\bf Summary of sections 3.1, 3, 4:}  
Solve linear homogeneous 2nd order DE with constant coefficients.

Solve \(ay'' + by' + cy = 0\).~~ Educated guess \(y = e^{rt}\), then 

\(ar^2e^{rt} + bre^{rt} + ce^{rt} = 0\) implies \(ar^2 + br + c = 0\),
 
Suppose \(r = r_1, r_2\) are solutions to \(ar^2 + br + c = 0\)

\(r_1, r_2 = {-b \pm  qrt{b^2 - 4ac \over 2a}\)}

If \(r_1 \not= r_2\), then \(b^2 - 4ac \not= 0\).  Hence a general solution is
{\(y = c_1e^{r_1t} + c_2e^{r_2t}\)}

If \(b^2 - 4ac > 0\), general solution is \(y = c_1e^{r_1t} + c_2e^{r_2t}\).

If \(b^2 - 4ac < 0\), change format to linear combination of real-valued functions instead of 
complex valued functions by using Euler's formula.

general solution is \(y = c_1 e^{dt} cos (nt) + c_2 e^{dt} sin (nt)\) where \(r = d \pm in\)

If \(b^2 - 4ac = 0\), \(r_1 = r_2\), so need 2nd (independent) solution:  \(te^{r_1t}\)

Hence general solution is \(y = c_1e^{r_1t} + c_2te^{r_1t}\).

Initial value problem:  use \(y(t_0) = y_0\), \(y'(t_0) = y_0'\) to solve for \(c_1, c_2\) to find unique 
solution. 

\end

START 100/FALL16/e1_Fall2016ANS.tex part 7

\nopagenumbers

[15]~ 4a.)  Solve \(y'' - 8y' + 16y = 0\)
 fill

Educated guess:  \(y = e^{rt}\).  Then \(y' = re^{rt}\) and  \(y'' = r^2e^{rt}\)

Plugging in the guess into our equation:  

 \( r^2e^{rt} - 8 re^{rt} + 16e^{rt} = 0\)

Since  \(e^{rt} > 0\), we can divide both sides of the above equation by \( re^{rt}\) without loosing any solutions:

 \( r^2 - 8 r + 16 = 0\) implies \((r - 4)^2 = 0\) and thus \(r = 4\).
 
 Thus \(y = e^{4t}\) is a solution.  We can check by plugging in (as we did in class for a different example) that \(y = te^{4t}\) is also a solution.

Sidenote:  \(\{e^{4t}, te^{4t}\}\) is a linear independent set and thus a basis for our solution.  We can check linear independence by calculating the Wronskian.

 fill
Answer: \(\underline{~~y = c_1e^{4t + c_2te^{4t}~~}\)}
 fill
[15]~ 4b.)   Solve \(y'' - y' + 3y = 0\)
 fill
Educated guess:  \(y = e^{rt}\).  Then \(y' = re^{rt}\) and  \(y'' = r^2e^{rt}\)

Plugging in the guess into our equation: 

\( r^2e^{rt} -  re^{rt} + 3e^{rt} = 0\)

Since  \(e^{rt} > 0\), we can divide both sides of the above equation by \( re^{rt}\) without loosing any solutions:

 \( r^2 -  r + 3 = 0\) implies  $r = {1 \pm  qrt{1 - 4(3)} \over 2}
  = {1 \pm  qrt{11} \over 2} =  {1 \pm i qrt{11} \over 2}
 $.

 fill
Answer: \(\underline{~~y = c_1e^{t \over 2 cos({ qrt{11} \over 2}t) + c_2e^{t \over 2} sin({ qrt{11} \over 2}t)~~}\)}
 fill

\end

START 100/SPRING13/e1_2013ANS.txt part 4

5.)   Find the general solutions for the following three  differential equations.

[15]~  5B.)  \(y'' + 6y' + 9y = 0\)

\(r^2 + 6r + 9 = 0\)

\((r + 3)^2 = 0\) implies \(r = -3\)

 fill  fill
Answer 5B) \(\underline{~~y = c_1 e^{-3t + c_2 te^{-3t}~~}\)}
 fill  fill

\end

START 100/3_5.tex

Thm:  Suppose \(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

If \(\psi\) is a solution to

\(ay'' + by' + cy = g(t)\) [*],
Then \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is also a solution to [*].

Moreover if \(\gamma\) is also a solution to [*], then there exist constants \(c_1, c_2\) such that 

\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\)

Or in other words, \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to [*].

Proof:  

Define   \(L(f) = af'' + bf' + cf\).

Recall \(L\) is a linear function.

 Let \(h = c_1 \phi_1(t) + c_2 \phi_2(t)\).  Since \(h\) is a solution to the differential 
equation, \(ay'' + by' + cy = 0\),

Since \(\psi\) is a solution to 
 \(ay'' + by' + cy = g(t)\),

We will now show that \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h\) is also a solution to 
[*].

Since \(\gamma\) a solution to  \(ay'' + by' + cy = g(t)\),

We will first show that \(\gamma - \psi\) is a solution to the differential equation
 \(ay'' + by' + cy = 0\).

 fill

Since \(\gamma - \psi\) is a solution to \(ay'' + by' + cy = 0\) and 

\(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

there exist constants \(c_1, c_2\) such that

\(\gamma - \psi =\underline{\hskip 2in\)}

Thus
\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\).

Thm: \hfil \break
 Suppose  
\(f_1\) is a a solution to
\(ay'' + by' + cy = g_1(t)\)  \hfil \break
and \(f_2\) is a a solution to
\(ay'' + by' + cy = g_2(t)\), then \(f_1 + f_2\) is a solution to
\(ay'' + by' + cy = g_1(t) + g_2(t)\)

Proof: Let  \(L(f) = af'' + bf' + cf\).

Since \(f_1\) is a solution to 
 \(ay'' + by' + cy = g_1(t)\),

Since \(f_2\) is a solution to 
 \(ay'' + by' + cy = g_2(t)\),

We will now show that \(f_1 + f_2\) is a solution to \break
 \(ay'' + by' + cy = g_1(t) + g_2(t)\).

 fill

Sidenote:  The proofs above work even if \(a, b, c\) are functions of \(t\) instead of constants.

{\bf Examples:} Find a suitable form for \(\psi\) for the following differential equations:

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 fill

2.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 fill

3.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 fill

4.)  \(y'' - 5y  = 4sin(3t)\)
 fill

5.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 fill

6.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 fill

7.)  
\(y'' - 4y' - 5y = 4 sin(3t) e^{2t}\)
 fill
8.)  
\(y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}\)
 fill

9.) \(y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}\)
 fill
10.)  \(y'' - 4y' - 5y\)

 fill
11.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 fill
12.) \(y'' - 4y' - 5y = 4e^{-t}\)
 fill
To solve \(ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)\) [**]

1.)  Find the general solution to \(ay'' + by' + cy = 0\):  

\(c_1\phi_1 + c_2\phi_2\)

2.)  For each \(g_i\), find a solution to \(ay'' + by' + cy = g_i\):  

\(\psi_i\)

This includes plugging guessed solution {\(\psi_i\)} into \break
\(ay'' + by' + cy = g_i\).

The general solution to [**] is

\(c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n\)

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.) {\bf First solve homogeneous equation:}

  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to \hb
  LINEAR HOMOGENEOUS equation.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.) {\bf Find one solution to non-homogeneous eq'n:}
  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)
~~~~~~~~\(y' = 3A cos(3t) - 3B sin(3t)\)
~~~~~~~~\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.) {\bf If initial value problem:}
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

\((e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0\) and 
\((e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0\) 

\end
Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/3_5exA.tex

\font\frm=cmr9

3.5:  Solving 2nd order linear non-homogeneous DE using method of undetermined coefficients.
 
Example:  Solve  \(y'' + 4y = 12t + 8sin(2t)\).
---

 
Step 1:  Solve homogeneous system,  \(y'' + 4y = 0\)
 
\(r^2 + 4 = 0 ~~\Rightarrow~~ r^2 = -4 ~~\Rightarrow~~ r = \pm  qrt{-4} = 0 \pm 2i\)
 
Hence homogeneous soln is \(y = c_1cos(2t) + c_2sin(2t)\)

---

Step 2a:  Find one solution to  \(y'' + 4y = 12t\)
 
Possible guess:  \(y = At + B\).  Then \(y' = A\) and \(y'' = 0\).
 
Plug in: \( 0 + 4(At +B) = 12t\)
\(\Rightarrow\)
\(4At + 4B = 12t + 0\)   
 
Thus \(4A = 12\) and \(4B = 0 ~~\Rightarrow~~ A = 3\) and \(B = 0\)

Thus \(y = 3t\) is a solution to \(y'' + 4y = 12t\).

Simpler guess:  since there is no \(y'\) term, we didn't need the B term in our guess.  We could have guessed \(y = At\) instead for this particular problem (and other analogous problems).  If you make similar observations when you do your HW, you can save time when you do comparable problems.

Step 2b:  Find one solution to  \(y'' + 4y = 8sin(2t)\)

Incorrect guess:  \(y = Asin(2t)\).  Then \(y' = 2Acos(2t)\) and \(y'' = -4Asin(2t)\).

Note:  since no \(y'\) term, did not include a \(Bcos(2t)\) term in guess.

Plug in:  \(-4Asin(2t) +4Asin(2t) = 8sin(2t)\).  
\u 
Thus \(0 = 8sin(2t)\).
\u
Thus equation has no solution for \(A\).  Hence guess is wrong.  

Note this guess is wrong because \(y = sin(2t)\) is a homogeneous solution.  This is why we always solve homogeneous equations first.  If a function is a solution to a homogeneous equation, then no constant multiple of that function can be a solution to a non-homogeneous solution since it is a homogeneous solution.  

If your normal guess is a homogeneous solution: 
 \u
 Multiply it by \(t\) 
\u
until it is no longer a homogeneous solution.

Incorrect guess:  \(y = Atsin(2t)\).  
\u
Then \(y' = Asin(2t)+ 2Atcos(2t)\) and
\u
 \(y'' = 2Acos(2t) + 2Acos(2t)- 4Atsin(2t)\)
 \u

Plug into  \(y'' + 4y = 8sin(2t)\):
\u
\( 4Acos(2t)- 4Atsin(2t) + 4Atsin(2t) = 8sin(2t)\)

\u
But this equation has no solution for \(A\).  Note we need to add a cosine term to our guess so that we can cancel out the cosine term on LHS:

---

Better guess:  \(y = t[Asin(2t) + Bcos(2t)]\).  %Then

---

Best guess:  \(y = Btcos(2t)\)

Then \(y' = Bcos(2t) - 2Btsin(2t)\)
\u
and \(y'' = -2Bsin(2t) - 2Bsin(2t) - 4Btcos(2t)  \)

Plug into  \(y'' + 4y = 8sin(2t)\)
\u
\( -4Bsin(2t) - 4Btcos(2t) + 4Btcos(2t) = 8sin(2t)\)
\u
\( -4Bsin(2t)  = 8sin(2t)  ~~\Rightarrow~~ -4B = 8  ~~\Rightarrow~~B = -2\)

Thus \(y = -2tcos(2t)\) is a solution to 

---

Note:  Guessing wrong is NOT a big deal.  You can use your wrong guess to determine a correct guess (though guessing right the first time will save you time).  

Recall you are looking for ONE solution to your NON-homogeneous equation. 

 \(\bullet\) If you find 
an infinite number of solns, choose one. 

  \(\bullet\) If your guess gives you one solution, use it.  

 \(\bullet\) If your guess leads to no solutions, than make a different (improved) educated guess.

---

To find general solution to non-homogeneous \hb LINEAR differential equation:  combine all solutions
 
\(y = c_1cos(2t) + c_2sin(2t) +3t -2tcos(2t)\)

\end

START 100/ch3.tex part 3

Thm:  Suppose \(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

If \(\psi\) is a solution to

\(ay'' + by' + cy = g(t)\) [*],
Then \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is also a solution to [*].

Moreover if \(\gamma\) is also a solution to [*], then there exist constants \(c_1, c_2\) such that 

\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\)

Or in other words, \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to [*].

Proof:  

Define   \(L(f) = af'' + bf' + cf\).

Recall \(L\) is a linear function.

 Let \(h = c_1 \phi_1(t) + c_2 \phi_2(t)\).  Since \(h\) is a solution to the differential 
equation, \(ay'' + by' + cy = 0\),

Since \(\psi\) is a solution to 
 \(ay'' + by' + cy = g(t)\),

We will now show that \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h\) is also a solution to 
[*].

Since \(\gamma\) a solution to  \(ay'' + by' + cy = g(t)\),

We will first show that \(\gamma - \psi\) is a solution to the differential equation
 \(ay'' + by' + cy = 0\).

 fill

Since \(\gamma - \psi\) is a solution to \(ay'' + by' + cy = 0\) and 

\(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

there exist constants \(c_1, c_2\) such that

\(\gamma - \psi =\underline{\hskip 2in\)}

Thus
\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\).

Thm: \hfil \break
 Suppose  
\(f_1\) is a a solution to
\(ay'' + by' + cy = g_1(t)\)  \hfil \break
and \(f_2\) is a a solution to
\(ay'' + by' + cy = g_2(t)\), then \(f_1 + f_2\) is a solution to
\(ay'' + by' + cy = g_1(t) + g_2(t)\)

Proof: Let  \(L(f) = af'' + bf' + cf\).

Since \(f_1\) is a solution to 
 \(ay'' + by' + cy = g_1(t)\),

Since \(f_2\) is a solution to 
 \(ay'' + by' + cy = g_2(t)\),

We will now show that \(f_1 + f_2\) is a solution to \break
 \(ay'' + by' + cy = g_1(t) + g_2(t)\).

 fill

Sidenote:  The proofs above work even if \(a, b, c\) are functions of \(t\) instead of constants.

{\bf Examples:} Find a suitable form for \(\psi\) for the following differential equations:

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 fill

2.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 fill

3.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 fill

4.)  \(y'' - 5y  = 4sin(3t)\)
 fill

5.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 fill

6.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 fill

7.)  
\(y'' - 4y' - 5y = 4 sin(3t) e^{2t}\)
 fill
8.)  
\(y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}\)
 fill

9.) \(y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}\)
 fill
10.)  \(y'' - 4y' - 5y\)

 fill
11.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 fill
12.) \(y'' - 4y' - 5y = 4e^{-t}\)
 fill
To solve \(ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)\) [**]

1.)  Find the general solution to \(ay'' + by' + cy = 0\):  

\(c_1\phi_1 + c_2\phi_2\)

2.)  For each \(g_i\), find a solution to \(ay'' + by' + cy = g_i\):  

\(\psi_i\)

This includes plugging guessed solution {\(\psi_i\)} into \break
\(ay'' + by' + cy = g_i\).

The general solution to [**] is

\(c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n\)

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.) {\bf First solve homogeneous equation:}

  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to \hb
  LINEAR HOMOGENEOUS equation.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.) {\bf Find one solution to non-homogeneous eq'n:}
  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)
~~~~~~~~\(y' = 3A cos(3t) - 3B sin(3t)\)
~~~~~~~~\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.) {\bf If initial value problem:}
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

\((e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0\) and 
\((e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0\) 

Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

START 100/ch3_134.tex part 5 AND 100/ch3new.tex part 5

{\bf Examples:} Find a suitable form for \(\psi\) for the following differential equations:

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 fill

2.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 fill

3.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 fill

4.)  \(y'' - 5y  = 4sin(3t)\)
 fill

5.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 fill

6.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 fill

7.)  
\(y'' - 4y' - 5y = 4 sin(3t) e^{2t}\)
 fill
8.)  
\(y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}\)
 fill

9.) \(y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}\)
 fill
10.)  \(y'' - 4y' - 5y\)

 fill
11.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 fill
12.) \(y'' - 4y' - 5y = 4e^{-t}\)
 fill
To solve \(ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)\) [**]

1.)  Find the general solution to \(ay'' + by' + cy = 0\):  

\(c_1\phi_1 + c_2\phi_2\)

2.)  For each \(g_i\), find a solution to \(ay'' + by' + cy = g_i\):  

\(\psi_i\)

This includes plugging guessed solution {\(\psi_i\)} into \break
\(ay'' + by' + cy = g_i\).

The general solution to [**] is

\(c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n\)

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.) {\bf First solve homogeneous equation:}

  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to \hb
  LINEAR HOMOGENEOUS equation.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.) {\bf Find one solution to non-homogeneous eq'n:}
  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)
~~~~~~~~\(y' = 3A cos(3t) - 3B sin(3t)\)
~~~~~~~~\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.) {\bf If initial value problem:}
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

\((e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0\) and 
\((e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0\) 

Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

START 100/3_5.tex

Thm:  Suppose \(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

If \(\psi\) is a solution to

\(ay'' + by' + cy = g(t)\) [*],
Then \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is also a solution to [*].

Moreover if \(\gamma\) is also a solution to [*], then there exist constants \(c_1, c_2\) such that 

\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\)

Or in other words, \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to [*].

Proof:  

Define   \(L(f) = af'' + bf' + cf\).

Recall \(L\) is a linear function.

 Let \(h = c_1 \phi_1(t) + c_2 \phi_2(t)\).  Since \(h\) is a solution to the differential 
equation, \(ay'' + by' + cy = 0\),

Since \(\psi\) is a solution to 
 \(ay'' + by' + cy = g(t)\),

We will now show that \(\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h\) is also a solution to 
[*].

Since \(\gamma\) a solution to  \(ay'' + by' + cy = g(t)\),

We will first show that \(\gamma - \psi\) is a solution to the differential equation
 \(ay'' + by' + cy = 0\).

 fill

Since \(\gamma - \psi\) is a solution to \(ay'' + by' + cy = 0\) and 

\(c_1 \phi_1(t) + c_2 \phi_2(t)\) is a general solution to

\(ay'' + by' + cy = 0\),

there exist constants \(c_1, c_2\) such that

\(\gamma - \psi =\underline{\hskip 2in\)}

Thus
\(\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)\).

Thm: \hfil \break
 Suppose  
\(f_1\) is a a solution to
\(ay'' + by' + cy = g_1(t)\)  \hfil \break
and \(f_2\) is a a solution to
\(ay'' + by' + cy = g_2(t)\), then \(f_1 + f_2\) is a solution to
\(ay'' + by' + cy = g_1(t) + g_2(t)\)

Proof: Let  \(L(f) = af'' + bf' + cf\).

Since \(f_1\) is a solution to 
 \(ay'' + by' + cy = g_1(t)\),

Since \(f_2\) is a solution to 
 \(ay'' + by' + cy = g_2(t)\),

We will now show that \(f_1 + f_2\) is a solution to \break
 \(ay'' + by' + cy = g_1(t) + g_2(t)\).

 fill

Sidenote:  The proofs above work even if \(a, b, c\) are functions of \(t\) instead of constants.

{\bf Examples:} Find a suitable form for \(\psi\) for the following differential equations:

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 fill

2.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 fill

3.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 fill

4.)  \(y'' - 5y  = 4sin(3t)\)
 fill

5.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 fill

6.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 fill

7.)  
\(y'' - 4y' - 5y = 4 sin(3t) e^{2t}\)
 fill
8.)  
\(y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}\)
 fill

9.) \(y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}\)
 fill
10.)  \(y'' - 4y' - 5y\)

 fill
11.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 fill
12.) \(y'' - 4y' - 5y = 4e^{-t}\)
 fill
To solve \(ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)\) [**]

1.)  Find the general solution to \(ay'' + by' + cy = 0\):  

\(c_1\phi_1 + c_2\phi_2\)

2.)  For each \(g_i\), find a solution to \(ay'' + by' + cy = g_i\):  

\(\psi_i\)

This includes plugging guessed solution {\(\psi_i\)} into \break
\(ay'' + by' + cy = g_i\).

The general solution to [**] is

\(c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n\)

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.) {\bf First solve homogeneous equation:}

  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to \hb
  LINEAR HOMOGENEOUS equation.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.) {\bf Find one solution to non-homogeneous eq'n:}
  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)
~~~~~~~~\(y' = 3A cos(3t) - 3B sin(3t)\)
~~~~~~~~\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.) {\bf If initial value problem:}
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

\((e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0\) and 
\((e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0\) 

\end
Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/3_5ex.tex

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

1.)  Find the general solution to \(y'' - 4y' - 5y  = 0\):  

Guess \(y = e^{rt}\) for HOMOGENEOUS equation:

\(y' = re^{rt}\), \(y' = r^2e^{rt}\)

 \(y'' - 4y' - 5y = 0\)
 
\(  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0\)

\(  e^{rt}(r^2 -  4r - 5) = 0\)

\( e^{rt} \not= 0\), thus can divide both sides by \(e^{rt}\): 
\( r^2 -  4r - 5 = 0\)

\((r +1)(r  - 5) = 0\).  Thus \(r = -1, 5\). 

  Thus \(y = e^{-t}\) and \(y = e^{5t}\) are both solutions to  HOMOGENEOUS equation.

Thus the general solution to the 2nd order linear  HOMOGENEOUS equation is 

\(y = c_1e^{-t + c_2e^{5t}\)}

2.)  Find a solution to \(ay'' + by' + cy = 4sin(3t)\):  

Guess \(y = A sin(3t) + B cos(3t)\)

\(y' = 3A cos(3t) - 3B sin(3t)\)

\(y'' = -9A sin(3t)  - 9B cos(3t)\)

\(y'' - 4y' - 5y = 4sin(3t)\)

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
   \noalign{---ule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since \(\{sin(3t), cos(3t)\}\) is a linearly independent set:

\(12B - 14A = 4\) and \(-14B - 12A = 0\)

Thus \(A = -{14 \over 12}B = -{7 \over 6}B\) and 

\(12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4\)

Thus \(B = 4({3 \over 85 }) = {12 \over 85}\) and $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus \(y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)\) is one solution to the nonhomogeneous equation.

Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\(y = c_1e^{-t + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)}

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find \(c_1, c_2\)).}

NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to satisfy the initial values.

Solve \(y'' - 4y' - 5y = 4sin(3t)\), ~\(y(0) = 6\), \(y'(0) = 7\).

General solution:
\(y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\)

Thus \(y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)\)

\(y(0) = 6\): ~~~~ \(6 =  c_1 + c_2  + {12 \over 85}\)
~~~~~~~~~ \( {498 \over 85} =  c_1 + c_2 \)

\(y'(0) = 7\):  ~~~~\(7 =  -c_1 + 5c_2 -{42 \over 85}\)
 ~~~~~~~\({637 \over 85} =  -c_1 + 5c_2 \)

\(6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}\).
Thus \(c_2 = {227 \over 102}\).

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

Thus \(y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)\).

 fill
Partial Check:  \(y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6\).

~~~~~~~~~~~~~~~~~~~~\(y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7\).

Potential proofs for exam 1:

Proof by (counter) example:

\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

Prove convergence of a series using ratio test.

Induction proof.

Prove a function is linear. 

Theorem 3.2.2:  If \(y = \phi_1(t)\) and \(y = \phi_2(t)\) are solutions to the 2nd order linear ODE, \(ay'' + by' + cy = 0\), then their linear combination \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is also a solution for constants \(c_1\) and \(c_2\).

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end

\end

START 100/3_5exA.tex

\font\frm=cmr9

3.5:  Solving 2nd order linear non-homogeneous DE using method of undetermined coefficients.
 
Example:  Solve  \(y'' + 4y = 12t + 8sin(2t)\).
---

 
Step 1:  Solve homogeneous system,  \(y'' + 4y = 0\)
 
\(r^2 + 4 = 0 ~~\Rightarrow~~ r^2 = -4 ~~\Rightarrow~~ r = \pm  qrt{-4} = 0 \pm 2i\)
 
Hence homogeneous soln is \(y = c_1cos(2t) + c_2sin(2t)\)

---

Step 2a:  Find one solution to  \(y'' + 4y = 12t\)
 
Possible guess:  \(y = At + B\).  Then \(y' = A\) and \(y'' = 0\).
 
Plug in: \( 0 + 4(At +B) = 12t\)
\(\Rightarrow\)
\(4At + 4B = 12t + 0\)   
 
Thus \(4A = 12\) and \(4B = 0 ~~\Rightarrow~~ A = 3\) and \(B = 0\)

Thus \(y = 3t\) is a solution to \(y'' + 4y = 12t\).

Simpler guess:  since there is no \(y'\) term, we didn't need the B term in our guess.  We could have guessed \(y = At\) instead for this particular problem (and other analogous problems).  If you make similar observations when you do your HW, you can save time when you do comparable problems.

Step 2b:  Find one solution to  \(y'' + 4y = 8sin(2t)\)

Incorrect guess:  \(y = Asin(2t)\).  Then \(y' = 2Acos(2t)\) and \(y'' = -4Asin(2t)\).

Note:  since no \(y'\) term, did not include a \(Bcos(2t)\) term in guess.

Plug in:  \(-4Asin(2t) +4Asin(2t) = 8sin(2t)\).  
\u 
Thus \(0 = 8sin(2t)\).
\u
Thus equation has no solution for \(A\).  Hence guess is wrong.  

Note this guess is wrong because \(y = sin(2t)\) is a homogeneous solution.  This is why we always solve homogeneous equations first.  If a function is a solution to a homogeneous equation, then no constant multiple of that function can be a solution to a non-homogeneous solution since it is a homogeneous solution.  

If your normal guess is a homogeneous solution: 
 \u
 Multiply it by \(t\) 
\u
until it is no longer a homogeneous solution.

Incorrect guess:  \(y = Atsin(2t)\).  
\u
Then \(y' = Asin(2t)+ 2Atcos(2t)\) and
\u
 \(y'' = 2Acos(2t) + 2Acos(2t)- 4Atsin(2t)\)
 \u

Plug into  \(y'' + 4y = 8sin(2t)\):
\u
\( 4Acos(2t)- 4Atsin(2t) + 4Atsin(2t) = 8sin(2t)\)

\u
But this equation has no solution for \(A\).  Note we need to add a cosine term to our guess so that we can cancel out the cosine term on LHS:

---

Better guess:  \(y = t[Asin(2t) + Bcos(2t)]\).  %Then

---

Best guess:  \(y = Btcos(2t)\)

Then \(y' = Bcos(2t) - 2Btsin(2t)\)
\u
and \(y'' = -2Bsin(2t) - 2Bsin(2t) - 4Btcos(2t)  \)

Plug into  \(y'' + 4y = 8sin(2t)\)
\u
\( -4Bsin(2t) - 4Btcos(2t) + 4Btcos(2t) = 8sin(2t)\)
\u
\( -4Bsin(2t)  = 8sin(2t)  ~~\Rightarrow~~ -4B = 8  ~~\Rightarrow~~B = -2\)

Thus \(y = -2tcos(2t)\) is a solution to 

---

Note:  Guessing wrong is NOT a big deal.  You can use your wrong guess to determine a correct guess (though guessing right the first time will save you time).  

Recall you are looking for ONE solution to your NON-homogeneous equation. 

 \(\bullet\) If you find 
an infinite number of solns, choose one. 

  \(\bullet\) If your guess gives you one solution, use it.  

 \(\bullet\) If your guess leads to no solutions, than make a different (improved) educated guess.

---

To find general solution to non-homogeneous \hb LINEAR differential equation:  combine all solutions
 
\(y = c_1cos(2t) + c_2sin(2t) +3t -2tcos(2t)\)

\end

START 100/3_5listSans.tex

Guess a possible non-homog soln for the  following DEs:
\u

Note homogeneous solution to \(y'' - 4y' - 5y = 0\)~~ is~~ \(y = c_1e^{-t} + c_2e^{5t}\)
since \(r^2 - 4r - 5 = (r-5)(r+1) = 0\)

1.)  
\(y'' - 4y' - 5y = 4e^{2t}\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ Ae^{2t}~~~~}\)}

We want to plug in a solution for \(y\) into the left-hand side (LHS) of the equation that will give us the right-hand side (RHS) of the equation.  In this case we need the output of the LHS to be a multiple of \(e^{2t}\) (in particular the output of the LHS when pluggin in \(y\) needs to be the RHS which is \(4e^{2t}\)).  

Thus we guess \(y = Ae^{2t}\).  Plugging  this into the LHS, we can solve for \(A\) so that we get the RHS, \(4e^{2t}\), thus finding a non-homogeneous solution.

2a.)  \(y'' - 4y' - 5y = t^2 - 2t + 1\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ At^2 + Bt + C  ~~~~}\)}

2b.)  \(y'' - 4y' - 5y = t^2 \)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~  At^2 + Bt + C  ~~~~}\)}

2c.)  \(y'' - 4y' - 5y = \) a degree 2 polynomial
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~  At^2 + Bt + C   ~~~~}\)}

Note that the non-homog solution guess is the same for 2a, 2b, 2c.  In each case, we need to guess a solution such that when we plug it into the LHS we get the RHS, a degree 2 polynomial.  Thus our guess is a degree 2 polynomial (but compare this to example 12).  Note we need \(y ~=~  At^2 + Bt + C\) even in case 2b.  Make sure you understand why \(y = At^2\) won't work.

3a.)  \(y'' - 4y' - 5y = 30\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~A   ~~~~}\)}

We have a constant on the RHS, so I guess a constant (but again compare to example 12).  If you are observant, you may note that a non-homog solution is \(y = -6\))

4a.)  \(y'' - 4y' - 5y = 4sin(3t)\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ Asin(3t) + Bcos(3t)  ~~~~}\)}

4b.) \(y'' - 4y' - 5y =4sin(3t) +  5cos(3t)\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~   Asin(3t) + Bcos(3t) ~~~~}\)}

4c.) \(y'' - 4y' - 5y = 5cos(3t)\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~   Asin(3t) + Bcos(3t) ~~~~}\)}

Note that the non-homog solution guess is the same for 4a, 4b, 4c.   If we plug in \(y = Asin(3t)\), the output will contain both \(sin(3t)\) and \(cos(3t)\) terms.  Thus I need to include both these terms in my guess.  Compare to example 11.

5.) \(y'' - 4y' - 5y = 4e^{-t}\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ Ate^{-t}  ~~~~}\)}

Note \(y = Ae^{-t}\) is a homogeneous solution.  Thus if I plug in it, I will get 0.  But I want the RHS, \( 4e^{-t}\).  When a guess doesn't work because it is a homogeneous solution, multiple by \(t\).  

Sidenote:  this trick works because when you plug it in, you must use the product rule; the  homogeneous part \(e^{-t}\) of \(y = Ate^{-t}\) will result in a number of cancellations, but the \(t\) part will give you terms that don't cancel out and whose sum is the RHS.

Observe \(y =  Ate^{-t}\), ~~\(y' =  Ae^{-t} - Ate^{-t}\), ~~
\(y'' =  -Ae^{-t} - Ae^{-t} + Ate^{-t} = -2 Ae^{-t} + Ate^{-t}\)

Thus $y'' - 4y' - 5y = 
- 2Ae^{-t} + Ate^{-t} - 4( Ae^{-t} - Ate^{-t}) - 5(Ate^{-t})$

\(= - 2Ae^{-t}  - 4 Ae^{-t} + At(e^{-t} + 4 e^{-t} - 5e^{-t})\)
\(= - 6 Ae^{-t} + At(0)\)
\(= 4e^{-t}\) when \(A = -{2 \over 3}\)
DO NOT FORGET THE PRODUCT RULE!!!!

6.)   \(y'' - 4y' - 5y =(e^t) + (e^{-t}) + (2t^3 + 3t^2) + (4sin(3t) +  5cos(3t))\)
Guess: \(\underline{~~y ~=~ (A_1e^{t) + (A_2te^{-t}) + (A_3t^3 + B_3t^2 + C_3t + D_3) + (A_4sin(3t) + B_4cos(3t))   ~~~~}\)}

Note if we wanted to find a non-homogeneous solution, we would need to determine all our undetermined coefficients.  Note we have 8 undetermined coefficients.  Instead of solving for them all at once (which would require 8 equations for the 8 unknowns), it is easier to divide finding a non-homogeneous solution into 4 simpler parts indicated by the parenthesis and subscripts as described below:  

a.)  Find \(A_1\) by plugging \(y = A_1e^{t}\) into \(y'' - 4y' - 5y =e^t \)

b.)  Find \(A_2\) by plugging \(y = A_2te^{-t}\) into \(y'' - 4y' - 5y =e^{-t} \)

c.) Find \(A_3, B_3, C_3, D_3\) by plugging \(y = A_3t^3 + B_3t^2 + C_3t + D_3\) into \(y'' - 4y' - 5y = 2t^3 + 3t^2 \)

d.) Find \(A_4, B_4\) by plugging \(y = A_4sin(3t) + B_4cos(3t)\) into 

We get the non-homogeneous solution by adding together the non-homogeneous solutions obtained from the above 4 parts since our diff eqn is LINEAR.

We get the general solution by combining the general homogeneous solution with this non-homogeneous solution.

7.)   \(y'' - 4y' - 5y =e^t + e^{-t} + 2t^3 + 3t^2 + 4sin(3t) +  5cos(t)\)

\hskip 0.15in{Guess: \(\underline{~~y ~=~ (A_1e^{t}) + (A_2te^{-t}) + (A_3t^3 + B_3t^2 + C_3t + D_3)~~ }\)}

8.)  
\(y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ (At^2 + Bt + C)e^{2t}   ~~~~}\)}

Since the RHS is a product, we guess a product.   \hb
Note I could have guessed
\(y ~=~ (At^2 + Bt + C)De^{2t} ~=~ (ADt^2 + BDt + CD)e^{2t}\),
but since \(AD, BD, CD\) are just constants, I don't need \(D\).

---  skip -5pt ---  skip -5pt ---  skip 0pt

Note homogeneous solution to \(y'' - 6y' +9y = 0\) ~~is~~ \(y = c_1e^{3t} + c_2te^{3t}\)
\o
\hskip 1.55in{since \(r^2 - 6r +9 = (r-3)(r-3) = 0\)}

9.)  \(y'' - 6y' +9y = 7e^{3t}\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ At^2e^{3t}  ~~~~}\)}
\o
Note neither \(y = Ae^{3t}\) nor \(y = Ate^{3t}\) will work since both are homogeneous solutions.  But our trick of multiplying by \(t\) until we have a guess that is not a homogeneous solution will work.

10.)  \(y'' - 6y' +9y = 7e^{-3t}\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~ Ae^{-3t}  ~~~~}\)}
\o
\(y ~=~ Ae^{-3t}\) is not a homogeneous solution (when \(A \not= 0\)).
---  skip -5pt ---  skip -5pt ---  skip 0pt
Some special cases:

11.)  \(y'' - 5y  = 4sin(3t)\)
 
\hskip 0.5in{Best Guess: \(\underline{~~~~y ~=~ Asin(3t)  ~~~~}\)}
\o
Note, we also could have guessed \(y = Asin(3t) + Bcos(3t)\), but since there is no \(y'\) term, we don't need the cosine term.  But both guesses will work.  Plugging in \(y = Asin(3t) + Bcos(3t)\) will take a little more work, but you will still get the right answer.

12.)  \(y'' - 4y'  = t^2 - 2t + 1\)
 
\hskip 0.5in{Guess: \(\underline{~~~~y ~=~At^3 + Bt^2 +Ct   ~~~~}\)}

\o
Note there is no \(y\) term on the LHS.  Thus to get a \(t^2\) term when we plug in our guess, we will need to plug in a \(t^3\) term. Hence we guess a degree 3 polynomial.  Note we don't need to include a constant term;  we could have guessed \(y ~=~At^3 + Bt^2 +Ct + D\), but any constant \(D\) will work (and hence there are an infinite number of solutions for \(D\)) so we might as well take \(D = 0\).

---

Don't worry too much about guessing wrong.  You will usually be able to figure out why an incorrect guess doesn't work and use that info to determine a better guess.

\end

START 100/FALL16/quiz3Fall16Answers.tex

\nopagenumbers

Find the solution to the initial value problem:
\(y'' - 6y' + 9y = 8e^{3t + 27t\),~~ \(y(0) = 5\),~~ \(y'(0) = 2\).}

---

Step 1:  Solve Homogeneous equation \(y'' - 6y' + 9y = 0\)

Let \(y = e^{rt}\).  Then \(r^2 - 6r + 9 = 0\).  Thus \((r - 3)^2 = 0\) and \(r = 3\).

Thus general homogeneous solution is \(y = c_1e^{3t} + c_2te^{3t}\).

---

Step 2a:   Solve Non-homogeneous equation \(y'' - 6y' + 9y = 8e^{3t} \)

Since \(y = e^{3t}\) and \(y = te^{3t}\) are homogeneous solutions, multiples of these cannot be solutions to the non-homogeneouse equation.  Thus we will try multiplying by another \(t\) and try  \(y = At^2e^{3t}\).  

\(y = At^2e^{3t}\) implies \(y' = 2Ate^{3t} +  3At^2e^{3t}\) and   
\(y'' = 2Ae^{3t} +  6Ate^{3t} +  6Ate^{3t} +  9At^2e^{3t}\)

Plugging into \(y'' - 6y' + 9y = 8e^{3t} \) and solve for \(A\):

\(2Ae^{3t} +  12Ate^{3t} +  9At^2e^{3t} - 6(2Ate^{3t} +  3At^2e^{3t}) + 9(At^2e^{3t}) ~=~ 8e^{3t}\)

\(2Ae^{3t} +  12Ate^{3t} +  9At^2e^{3t} - 12Ate^{3t} -  18At^2e^{3t} + 9At^2e^{3t} ~=~ 8e^{3t}\)

\(2Ae^{3t} +  (12 - 12)Ate^{3t} +  (9 - 18 + 9)At^2e^{3t}  ~=~ 8e^{3t}\)

\(2Ae^{3t}   ~=~ 8e^{3t}\) implies \(2A ~=~ 8\) and thus \(A = 4\).

Thus \(y = 4t^2e^{3t}\) is anon-homogeneous solution to  \(y'' - 6y' + 9y = 8e^{3t}\). 

Thus  general non-homogeneous solution to  \(y'' - 6y' + 9y = 8e^{3t} \) is 
  skip 5pt
 \(y = c_1e^{3t + c_2te^{3t} + 4t^2e^{3t}\).}
---
Step 2b:   Solve Non-homogeneous equation \(y'' - 6y' + 9y = 27t\)

Guess \(y = At + B\).  Then \(y' = A\) and \(y'' = 0\).

Plugging into \(y'' - 6y' + 9y = 27t \) and solve for \(A\) and \(B\):

 \(0 - 6A + 9(At + B) = 27t \)

 \(9At + 9B - 6A = 27t + 0 \).  Thus \(9A = 27\) and  \(9B - 6A = 0\).
 
 Hence \(A = 3\) and \(9B = 6A = 6(3)\) and thus \(B = 2\).
 
 Thus \(y = 3t + 2\) is a non-homogeneous solution to  \(y'' - 6y' + 9y = 27t\). 
 
 Thus  general non-homogeneous solution to  \(y'' - 6y' + 9y = 27t \) is 

   skip 5pt
 \(y = c_1e^{3t + c_2te^{3t} + 3t + 2\).}
 ---
 
 {\bf Hence  general non-homogeneous solution to  \(y'' - 6y' + 9y = 8e^{3t} + 27t\) is }
  skip 5pt
 \(y = c_1e^{3t + c_2te^{3t} + 4t^2e^{3t} + 3t + 2\).}
  
  or equivalently,
  
   \(y = e^{3t(4t^2 + c_2t +c_1) + 3t + 2\).}
  

  
  
  Step 3:  Use initial values to solve for \(c_1\) and \(c_2\):    

\(y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2\)
~~implies~~
\(y' = 3e^{3t}(4t^2 + c_2t +c_1) +  e^{3t}(8t + c_2) + 3\)

 \(y(0) = 5\): ~~ \(5 =  e^{0} (4(0)^2 + c_2(0) +c_1)  + 3(0) + 2\)
 
 \hskip 0.77in \(5 =  c_1  + 2\) implies \(c_1 = 3\)
 
   \(y'(0) = 2\):  ~~ \(2 = 3e^{0} (4(0)^2 + c_2(0) +c_1) +  e^{0}(8(0) + c_2) + 3\)

 \hskip 0.78in \(2 =  3c_1  + c_2 + 3\) implies \(c_2 = 2 - 3 - 3c_1 = 2 - 3 - 9 = -10\)
 
 Thus solution to IVP is \(y = e^{3t}(4t^2 -10t +3) + 3t + 2\).
 
 
---  skip -5pt ---
Quiz 3  \hfil  
Oct. 14, 2016 

1.)  Suppose \(y = c_1 e^{3t} + c_2te^{3t} + 4t^2e^{3t}\)
is a solution to \(y'' - 6y' + 9y = 8e^{3t}\).
Find the solution to the initial value problem:
\(y'' - 6y' + 9y = 8e^{3t + 27t\),~~ \(y(0) = 5\),~~ \(y'(0) = 2\).}

Note:  Solving this IVP is a 4 part problem, but I have already done the first two parts for you.

{\bf ANSWER:}  Since  \(y = c_1 e^{3t} + c_2te^{3t} + 4t^2e^{3t}\)
is a solution to \(y'' - 6y' + 9y = 8e^{3t}\), 
we know \(y = c_1 e^{3t} + c_2te^{3t}\) is the general solution to the homogeneous equation
is a solution to \(y'' - 6y' + 9y = 0\) ~~and~~
\(y = 4t^2e^{3t}\)
is a solution to \(y'' - 6y' + 9y = 8e^{3t}\).

Thus parts 1 and 2a are already completed.  Repeating the remaining 2 parts:

{\bf Step 2b:}   Solve Non-homogeneous equation \(y'' - 6y' + 9y = 27t\):~~
Guess \(y = At + B\).  Then \(y' = A\) and \(y'' = 0\).

Plugging into \(y'' - 6y' + 9y = 27t \) and solve for \(A\) and \(B\):
~~~~
 \(0 - 6A + 9(At + B) = 27t \)

 \(9At + 9B - 6A = 27t + 0 \).  Thus \(9A = 27\) and  \(9B - 6A = 0\).
 
 Hence \(A = 3\) and \(9B = 6A = 6(3)\) and thus \(B = 2\).
 
 Thus \(y = 3t + 2\) is a non-homogeneous solution to  \(y'' - 6y' + 9y = 27t\). 
 
 Thus  general non-homogeneous solution to  \(y'' - 6y' + 9y = 27t \) ~~is~~ 
{\(y = c_1e^{3t} + c_2te^{3t} + 3t + 2\).}
 ---
  skip -5pt
  Thus  general non-homog. soln to  \(y'' - 6y' + 9y = 8e^{3t} + 27t\) is
 % skip 5pt
 %\centerline
 {\(y = c_1e^{3t} + c_2te^{3t} + 4t^2e^{3t} + 3t + 2\).}
  
  or equivalently,
  %\centerline
   {\(y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2\).}
  ---

  
  
 {\bf Step 3:}  Use initial values to solve for \(c_1\) and \(c_2\):    

\(y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2\)
~~implies~~
\(y' = 3e^{3t}(4t^2 + c_2t +c_1) +  e^{3t}(8t + c_2) + 3\)

 \(y(0) = 5\): ~~ \(5 =  e^{0} (4(0)^2 + c_2(0) +c_1)  + 3(0) + 2\)
 %\hskip 0.77in
 implies
   \(5 =  c_1  + 2\) implies \(c_1 = 3\)
 
   \(y'(0) = 2\):  ~~ \(2 = 3e^{0} (4(0)^2 + c_2(0) +c_1) +  e^{0}(8(0) + c_2) + 3\)

 \hskip 0.78in 
\(2 =  3c_1  + c_2 + 3\) implies \(c_2 = 2 - 3 - 3c_1 = 2 - 3 - 9 = -10\)
 
 Thus solution to IVP is \(y = e^{3t}(4t^2 -10t +3) + 3t + 2\).
 
 fill

Answer:  \underbar{~~~ \(y = e^{3t(4t^2 -10t +3) + 3t + 2\)~~~}}

\end

START 100/FALL17/quiz4Fall2017formBans.tex part 2

3.)  Suppose that \(y_1(t) = t\) and \(y_2(t) = t^2\) are solutions to the differential equation, \(y'' + p(t)y' + q(t)y = 0\).  Find the general solution to   \(y'' + p(t)y' + q(t)y = {1 \over t}\)

General solution:  \(\underline{~y = c_1t + c_2t^2 -tln|t| ~}\)

\(W(t, t^2) = \left|\matrix{t  & t^2 \cr 1 & 2t} \right| = 2t^2 - t^2 = t^2\).  %~~~~~Check by Abel's theorem:  $W(t, t^2) = e^{-\int p_1(t) dt}

\({1 \over t}\left|\matrix{0 & t^2 \cr 1 & 2t} \right|  = \left|\matrix{0 & t^2 \cr {1 \over t} & 2t} \right| =-t\) \hfill 
\(u_1(t) =  \int {g(t) \over a}{W_1 \over W} = \int {-t \over t^2} dt = \int {-1 \over t} dt = - ln|t|\)

\(\left|\matrix{t  & 0 \cr 1 & {1 \over t}} \right| =1\)
\hfill 
\(u_2(t) =  \int {g(t) \over a}{W_2 \over W} = \int {1 \over t^2} dt = \int t^{-2} dt = -t^{-1}\)

Non-homog:\( -tln|t| - t^{-1}t^2 = -tln|t| - t\)

General solution: \(y = k_1t + c_2t^2 -tln|t| - t = (k_1- 1)t + c_2t^2 -tln|t|  = c_1t + c_2t^2 -tln|t|\)

FYI:  

 \(y'' + p(t)y' + q(t)y = 0\).

\(y_1(t) = t\), \(y' = 1\), \(y'' = 0\)

and \(y_2(t) = t^2\), \(y' = 2t\), \(y'' = 2\)

 \(0 + p(t) + q(t)t = 0\).

 \(2 + 2p(t)t + q(t)t^2 = 0\).

 \(0 + p(t)t + q(t)t^2 = 0\).

\(2 + tp(t) = 0\).  Thus \(p(t) = -{2 \over t}\)

\(0 +  -{2 \over t} + q(t)t = 0\).  Thus \(q(t) = {2 \over t^2}\)

 \(y'' -{2 \over t}y' +  {2 \over t^2}y = 0\).

By Abel's thm, \(W(t, t^2) = e^{\int{2 \over t} dt} = e^{2ln|t|} = e^{ln|t|^2} = t^2\) 

\end

START 100/FALL18/quiz3_2018ans.tex part 1

\nopagenumbers

Quiz 3 ~~~SHOW ALL WORK\hfill Oct 12, 2018

1.)  The solution to \(y'' + 16y =  36cos(2t) \) ~is~ \(y = c_1cos(4t) +  c_2sin(4t) + 3cos(2t)\)
\hfil \break
Use this fact to answer the following two questions.

[5]~ 1a.)  Guess a possible non-homog soln for the following differential equation 
(do
\hfill\break
 not solve):  ~~ \(y'' + 16y = 3sin(4t) -  e^{4t} \)

Guess \(y = t[Asin(4t) + Bcos(4t)] + Ce^{4t}\)

For explanations and examples, \hfill\break
see http://homepage.divms.uiowa.edu/\( im\)idarcy/COURSES/100/3\_5listSans.pdf
\hfill\break
and http://homepage.divms.uiowa.edu/\( im\)idarcy/COURSES/100/3\_5exA.pdf

 fill
 

[3]~ 1b.)  The general solution to  ~\(y'' + 16y = 36cos(2t) +  32\) ~is

 \(y = c_1cos(4t) +  c_2sin(4t) + 3cos(2t) + 2\)

See http://homepage.divms.uiowa.edu/\( im\)idarcy/COURSES/100/FALL18/18\_10\_15.pdf
 fill

\end

START 100/FALL16/e1_Fall2016ANS.tex part 8

[15]~  5.)  Let \(y = y_1(t)\) be a solution of \(y' + p(t)y = 0\) and let \(y = y_2(t)\) be a solution of \(y' + p(t)y = g(t)\).  Show that  \(y = y_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\).

Proof: Since \(y = y_1(t)\) is a solution of \(y' + p(t)y = 0\), we know that  
\(y_1' + p(t)y_1 = 0\).

Since \(y = y_2(t)\) is a solution of \(y' + p(t)y = g(t)\), ~~ 
\(y_2' + p(t)y_2 = g(t)\)

Claim:  \(y = y_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\).

 We will plug \(y = y_1(t) + y_2(t)\) into the LHS to determine that the LHS = RHS:
 
 \((y_1(t) + y_2(t))' + p(t)(y_1(t) + y_2(t)) =\)
\( y_1'(t) + y_2'(t) + p(t)y_1(t) + p(t)y_2(t) \)
 
 Hence \(y = y_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\).
  skip 10pt
 ---ule
 
 Alternate proof:  
  Since \(y = y_1(t)\) is a solution of \(y' + p(t)y = 0\), we know that  
  skip 10pt
\(y_1' + p(t)y_1 = 0 ~~~(1)\).

Since \(y = y_2(t)\) is a solution of \(y' + p(t)y = g(t)\).  
  skip 10pt
\(y_2' + p(t)y_2 = g(t) ~~~(2)\).

If we add equations (1) and (2), we obtain:

\][y_1'(t) + p(t)y_1(t)] + [y_2'(t) + p(t)y_2(t)]  = 0 + g(t)\[

Thus \( y_1'(t) + y_2'(t) + p(t)y_1(t) + p(t)y_2(t)  = g(t)\)

and  \((y_1(t) + y_2(t))' + p(t)(y_1(t) + y_2(t)) =g(t)\)

Hence \(y = y_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\).
  skip 10pt
 % skip -30pt
 ---ule
 
 Alternate proof: 
   skip -10pt
 Claim:  \(L(f) = f' + pf \) is a linear function where \(f\) and \(p\) are functions of \(t\).
 
 Proof of claim:  Let \(a, b\) be constants and \(f, g\) be functions of \(t\).
 
 $L(af + bg) = (af+ bg)' + p(af + bg)
 = af' + bg' + paf + pbg = af' + paf + bg' + pbg = 
 [a(f' + pf)] + [b(g' + pg)] = 
 L(f) + L(g)$
 
 
 We will now show that  \(y = y_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\):
 
 Since \(y = y_1(t)\) is a solution of \(y' + p(t)y = 0\), ~~\(L(y_1) = 0\).
 
  Since \(y = y_2(t)\) is a solution of \(y' + p(t)y = g(t)\),~~\(L(y_2) = g(t)\)
  
  \(L(y_1 + y_2) = L(y_1) + L(y_2) = 0 + g(t) = g(t)\).  Thus 
 \(y = y_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\).
  skip 10pt
 ---ule
 
 Note similar proofs would show that  \(y = cy_1(t) + y_2(t)\) is a solution of \(y' + p(t)y = g(t)\) for any constant \(c\).
 
\end

START 34/FALL10/e2_34_F10ANS.txt part 1

\nopagenumbers

Math 34 Differential Equations Exam \#2
                               October 29, 2010 \hfill  SHOW ALL
WORK
~~~

[4]~~ 2.)  Circle T for True or F for False:
\u\u  skip -4pt
 Suppose \(y = f(t)\) is a solution to $3y'' 
+ 10y = cos(t)\(, \)y(0) = 0\(, \)y'(0) = 0$ and suppose 
\(y = g(t)\) is a solution to $3y'' 
+  10y = cos(t)\(, \)y(0) = 100\(, \)y'(0) = -200$.
For large values of \(t\), \(f(t) - g(t)\) is very small.
  \hfill {\bf Note:  no damping}
{ ~~~~~~~~~F}

 fill

\end

START 100/SPRING13/final3600ANS.tex part 3

\documentclass[12pt]{article}

 etlength{\topmargin}{-0.9in}
 etlength{\oddsidemargin}{-.250in}
 etlength{\textwidth}{7.0in}
 etlength{\textheight}{10in}
\pagestyle{empty}  %% To avoid page numbering
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{relsize}

\usepackage{amsmath}

\AppendGraphicsExtensions{.gif}
\DeclareGraphicsExtensions{.pdf,.png,.gif,.jpg}

\begin{document}

22M:100 (MATH:3600:0001) Final Exam \hfil \break
May 15, 2013 \hfill SHOW ALL STEPS \hfill

[20]~ 4.)  Solve:  \( y'' - 3y' - 4y = 8t\), \(y(0) = 5\), \(y'(0) = 3\)

\(r^2 - 3r - 4 = 0\).  
~~~~~~~~~
\((r - 4)(r + 1) = 0\). ~~~~ Thus \(r = 4, -1\).

general homogeneous solution:  \(y = c_1e^{4t} + c_2e^{-t}\)

Let \(y = At + B\).  Then \(y' = A\) and \(y'' = 0\)

\(-3A - 4(At + B) = 8t\)

\(A = -2\), \(6 - 4B = 0\).  Thus \(B = \frac{3}{2}\)

Thus \(\psi = -2t + \frac{3}{2}\) is a non-homogeneous solution.

General solution to non-homogeneous DE:   \(y = c_1e^{4t} + c_2e^{-t} - 2t + \frac{3}{2}\)

\(y' =  4c_1e^{4t} - c_2e^{-t} - 2\)

\(5 = c_1 + c_2 + \frac{3}{2}\)

\(3 = 4c_1 - c_2 - 2\)

Thus \(c_1 =  \frac{17}{10}\) and \(c_2 =  \frac{9}{5}\)

Answer: \underline{~~ \(y = \frac{17{10}e^{4t} +   \frac{9}{5}e^{-t} - 2t + \frac{3}{2}\)~~}}
---

\end

START 34/FALL03/finalexamANS.txt part 5

[4]~ 1b.)  Without using the LaPlace transform, find the general solution to the differential 
equation
\(y'' +  2y' + y = e^{-t}\)

Note \(c_1e^{-t}\) and  \(c_2te^{-t}\) are both solutions to the homogenous equation (without the 
initial values).  Hence they cannot be solutions to the non-homogeneous equation.

Thus guess \(y = At^2e^{-t}\).  Then \(y' = A(2te^{-t} - t^2e^{-t})\) \hfil \break
$y'' 
= A(2e^{-t} - 2te^{-t}  -  2te^{-t} + t^2e^{-t})
= A(2e^{-t} - 4te^{-t}  + t^2e^{-t})

From \(y'' +  2y' + y = e^{-t}\), we get

\(A(2e^{-t} - 4te^{-t}  + t^2e^{-t}) + 2A(2te^{-t} - t^2e^{-t}) + At^2e^{-t}= e^{-t}\)

\(A(2e^{-t} - 4te^{-t}  + t^2e^{-t} + 4te^{-t} - 2t^2e^{-t} + t^2e^{-t})= e^{-t}\)

\(A(2e^{-t})= e^{-t}\).  Hence \(2A = 1\) and \(A = {1 \over 2}\)

Thus \({1 \over 2} t^2e^{-t}\) is a solution to \(y'' +  2y' + y = e^{-t}\)

\u
Answer 1b.) \(\underline{ {1 \over 2 t^2e^{-t} + c_1e^{-t} + c_2te^{-t} }\)}
\w
\u
Note if you guessed the wrong solution, but did everything else correctly, you could still 
earn 3 - 4pts for this problem depending on whether and how well you explained how your 
answer could not have been correct and how to get the right answer.

[3]~ 1c.)  Without using the LaPlace transform, solve the following initial value problem:

\(y'' +  2y' + y = e^{-t, ~y(0) = 0, ~y'(0) = 4\)}

\(y = {1 \over 2} t^2e^{-t} + c_1e^{-t} + c_2te^{-t}\)

\(y(0) = 0:~~0 =  0 + c_1 + 0\).  Hence \(c_1 = 0\).

\(y' = [te^{-t} - {1 \over 2} t^2e^{-t}]  + c_2[e^{-t} -te^{-t}]\)

\(y'(0) = 4:~~4 = [0 - 0] + c_2[1-0] \).  Hence \(c_2 = 4\).
\u
Answer 1c.) \(\underline{y(t) = {1 \over 2 t^2e^{-t} + 4te^{-t}}\)}
\u
Note the above problem graded based upon your answer to 1b.  Hence you could get full credit 
for this problem even if your answer to 1b was wrong.  

Note also that although these are the same solutions for \(c_1\) and \(c_2\) as in 1a, this is 
not usually the case, so you can't just take those solutions for the coefficients in 1c.
\u

\end

START 34/FALL03/finalreviewANS2.txt part 3

Note the following review problems DO NOT cover all problem types which may appear on the 
final.  

Chapter 3:

9.)  Solve the following initial problems:
 
9d.)  \(y'' + 6y' + 8y = cos(t), ~y(0) = 0, ~y'(0) = 0 \)

Guess \(y = Acos(t) + Bsin(t) \) is a solution.

Then \(y' = -Asin(t) + Bcos(t)\) and \(y'' = -Acos(t) - Bsin(t)\) 

\(-Acos(t) - Bsin(t) + 6(-Asin(t) + Bcos(t)) + 8(Acos(t) + Bsin(t)) = cos(t)\)

\((-A + 6B + 8A) cos(t) + (-B -6A + 8B) sin(t) = cos(t)\)

\((6B + 7A) cos(t) + (-6A + 7B) sin(t) = cos(t) + 0 sin(t)\)

\(cos(t)\) and \(sin(t)\) are linearly independent functions, 

Hence \( (-6A + 7B) sin(t) =  0 sin(t)\).  Thus \(-6A + 7B = 0\)  so \(A = {7 \over 6}B \)

And \((6B + 7A) cos(t)  = cos(t)\).  Thus \(6B + 7A = 1\), so \(6B + 7({7 \over 6}B ) = 1\).  
\({36 + 49 \over 6}B = {85 \over 6}B = 1\)

Thus 
\(B = {6 \over 85} \) and \(A = {7 \over 85} \).

Check: $-{7 \over 85} cos(t) - {6 \over 85} sin(t) + 6(-{7 \over 85} sin(t) + 
{6 \over 85} cos(t)) + 8({7 \over 85} cos(t) + {6 \over 85} sin(t)) = 
(-{7 \over 85}  + {36 \over 85} + {56 \over 85})cos(t) + 
(- {6 \over 85} - {42 \over 85} + {48 \over 85} )sin(t) = cos(t)$

Thus general solution is 
\(y(t) = c_1e^{-2t} + c_2e^{-4t} +  {7 \over 85} cos(t) + {6 \over 85} sin(t) \)

Use initial values to find \(c_1, ~c_2\):

\(0 = c_1 + c_2 + {7 \over 85} + 0\).  Hence \(c_1 = -c_2 - {7 \over 85}\) 

\(y'(t) = -2c_1e^{-2t} + -4c_2e^{-4t} -  {7 \over 85} sin(t) + {6 \over 85} cos(t) \)

$0 = -2c_1 - 4c_2 - 0 + {6 \over 85} 
= -2(-c_2 - {7 \over 85} )- 4c_2 + {6 \over 85} 
= 2c_2 + {14 \over 85} - 4c_2 + {6 \over 85} 
= -2c_2 + {20 \over 85}$ 

Thus \(2c_2 = {20 \over 85}\) or \(c_2 = {10 \over 85} \) and 
\(c_1 = -{10 \over 85} - {7 \over 85} = -{17 \over 85}\) 

Note these are NOT the same values for \(c_1\) and \(c_2\) for the homogeneous case.

The the solution to the initial value problem is
$y(t) = -{17 \over 85}e^{-2t} + {10 \over 85} e^{-4t} 
+  {7 \over 85} cos(t) + {6 \over 85} sin(t) $

9e.)  \(y'' + 6y' + 9y = cos(t), ~y(0) = 0, ~y'(0) = 0 \)

Guess \(y = Acos(t) + Bsin(t) \) is a solution.

Then \(y' = -Asin(t) + Bcos(t)\) and \(y'' = -Acos(t) - Bsin(t)\) 

\(-Acos(t) - Bsin(t) + 6(-Asin(t) + Bcos(t)) + 9(Acos(t) + Bsin(t)) = cos(t)\)

\((-A + 6B + 9A) cos(t) + (-B -6A + 9B) sin(t) = cos(t)\)

\((6B + 8A) cos(t) + (-6A + 8B) sin(t) = cos(t) + 0 sin(t)\)

\(cos(t)\) and \(sin(t)\) are linearly independent functions, 

Hence \( (-6A + 8B) sin(t) =  0 sin(t)\).  Thus \(-6A + 8B = 0\)  so $A = {8 \over 6}B =
{4 \over 3}B  $

And \((6B + 8A) cos(t)  = cos(t)\).  Thus \(6B + 8A = 1\), so \(6B + 8({4 \over 3}B ) = 1\).  
\({18 + 32 \over 3}B = {50 \over 3}B = 1\)

Thus 
\(B = {3 \over 50} \) and \(A = {8 \over 100}  = {2 \over 25}\).

Thus general solution is 
\(y(t) = c_1e^{-3t} + c_2te^{-3t} +  {2 \over 25} cos(t) + {3 \over 50} sin(t) \)

Use initial values to find \(c_1, ~c_2\):

\(0 = c_1 + c_2(0) + {2 \over 25} + 0\).  Hence \(c_1 = - {2 \over 25}\) 

$y'(t) = -3c_1e^{-3t} + c_2e^{-3t} - 3c_2te^{-3t}  -  {2 \over 25} sin(t) 
+ {3 \over 50} cos(t) $

$0 = -3c_1 + c_2 - 0 - 0 + {3 \over 50} 
= -3(- {2 \over 25} ) + c_2 + {3 \over 50} 
= {6 \over 25}  + c_2 + {3 \over 50} 
= c_2 + {15 \over 50}$ 

Thus \(c_2 = -{15 \over 50} \) and 
\(c_1 = -{2 \over 25} \) 

Note these are NOT the same values for \(c_1\) and \(c_2\) for the homogeneous case.

The the solution to the initial value problem is

$y(t) = -{2 \over 25} e^{-3t}  -{15 \over 50} te^{-3t} + 
 {2 \over 25} cos(t) + {3 \over 50} sin(t) $

Check: \(y(0) = -{2 \over 25} + 0 + {2 \over 25} = 0\)

$y'(t) = {6 \over 25} e^{-3t} + {45 \over 50} te^{-3t} -{15 \over 50} e^{-3t}  
 - {2 \over 25} sin(t) + {3 \over 50} cos(t) $

$y'(0) = {6 \over 25}  + 0  -{15 \over 50}   
 - 0 + {3 \over 50} = {12 - 15 + 3 \over 50} = 0 $

Thus initial conditions are satisfied.  To fully check that this is the solution, 
should check if \(y'' + 6y' + 9y = cos(t)\) 

9f.)  \(y'' + 6y' + 10y = cos(t), ~y(0) = 0, ~y'(0) = 0\)

Hence general solution to the homogeneous equation is  
\(y(t) = c_1e^{-3t}sin(t) + c_2e^{-3t}cos(t)\)

Thus \(y = Acos(t) + Bsin(t) \) is not a solution to the homogeneous equation.

Guess \(y = Acos(t) + Bsin(t) \) is a solution.

Then \(y' = -Asin(t) + Bcos(t)\) and \(y'' = -Acos(t) - Bsin(t)\) 

\(-Acos(t) - Bsin(t) + 6(-Asin(t) + Bcos(t)) + 10(Acos(t) + Bsin(t)) = cos(t)\)

\((-A + 6B + 10A) cos(t) + (-B -6A + 10B) sin(t) = cos(t)\)

\((6B + 9A) cos(t) + (-6A + 9B) sin(t) = cos(t) + 0 sin(t)\)

\(cos(t)\) and \(sin(t)\) are linearly independent functions, 

Hence \( (-6A + 9B) sin(t) =  0 sin(t)\).  Thus \(-6A + 9B = 0\)  so $A = {9 \over 6}B =
{3 \over 2}B  $

And \((6B + 9A) cos(t)  = cos(t)\).  Thus \(6B + 9A = 1\), so \(6B + 9({3 \over 2}B ) = 1\).  
\({12 + 27 \over 2}B = {39 \over 2}B = 1\)

Thus 
\(B = {2 \over 39} \) and \(A = {3 \over 39}  = {1 \over 13}\).

Thus general solution is 
\(y(t) = c_1e^{-3t}sin(t) + c_2e^{-3t}cos(t) +  {2 \over 39} cos(t) + {3 \over 39} sin(t) \)

Use initial values to find \(c_1, ~c_2\):

\(0 = c_1(0) + c_2(1) + {2 \over 39} + 0\).  Hence \(c_2 = - {2 \over 39} \) 

$y'(t) =   -3c_1e^{-3t}sin(t) + c_1e^{-3t}cos(t) 
 -3c_2e^{-3t}cos(t) - c_2e^{-3t}sin(t)   
-  {2 \over 39} sin(t) 
+ {3 \over 39} cos(t) $

\(0 = 0 + c_1 -3c_2 + 0 - 0 + {3 \over 39} \)
Thus $c_1 = 3c_2-{3 \over 39}  = 3(- {2 \over 39}) -{3 \over 39} = {-12 \over 39}
{-4 \over 13} $  
 
Note these are NOT the same values for \(c_1\) and \(c_2\) for the homogeneous case.

The the solution to the initial value problem is

$y(t) = {-4 \over 13}e^{-3t}sin(t) + - {2 \over 39} e^{-3t}cos(t) +  
{2 \over 39} cos(t) + {3 \over 39} sin(t) $

Check: $y(0) = 0 - {2 \over 39}  +  
{2 \over 39}  + 0 = 0$

To fully check that this is the solution, 
should check if \(y'' + 6y' + 9y = cos(t)\) and if \(y'(0) = 0\).

\end

START 100/FALL16/quiz6F16ans.tex part 1

\nopagenumbers

Quiz 6 Math 2560 Sect 81\hfill  Show your work \break
Dec 2, 2016

[3]~~ 1.)  Circle T for True and F for false:

1a.)  If \(\phi\) is a solution to a first order {\bf linear homogeneous} differential equation, then \(c\phi\) is 
also a 
solution to this equation.
 \hfill  T~~~~~~~
  
1b.)   If \(\phi\) is a solution to a first order linear differential equation, then \(c\phi\) is also a 
solution to this equation.
 \hfill  ~~~~~~~F
 

\end

START 100/3_6.tex

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2' + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right] =
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/ch3 part 4

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/ch3_134.tex part 6

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/3_6.tex

3.6   Variation of Parameters~~~~~~\hfill
{Solve \(y'' - 2y' + y = e^t ln(t)\)}~~
{\bf 1)  Find homogeneous solutions:  Solve \(y'' - 2y' + y = 0\)}
\u
Guess:  \(y = e^{rt}\), then \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and
\(r^2e^{rt - 2re^{rt} + e^{rt} = 0\)
implies {\(r^2 - 2r + 1 = 0\)}}
\u
\((r - 1)^2 = 0\), and hence \(r = 1\)
    
General homogeneous solution:  \(y = c_1e^{t} + c_2 te^{t}\)
 
since have two linearly independent solutions:  \(\{e^{t}, te^{t}\}\) 
 

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\u

$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt \)

$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 \(= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt \)
 
 tln(t) - t$ }

$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

\(\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}\)
\hfill
\(\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}\)
     
General solution :
{ \(y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }
 
which simplifies to  \(y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t \)

Solve \(y'' + p(t)y' + q(t)y = g(t)\) where \(y = c_1\phi_1(t) + c_2\phi_2(t)\) is solution to homogeneous equation  \(y'' + p(t)y' + q(t)y = 0\)

Guess  \(y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)\)

\(y = u_1 \p_1 + u_2 \p_2\)
implies
\(y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2\)

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' + p(t)y' + q(t)y = g(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'\p_1 + u_2' \p_2 = 0\)}

 
\(y' = u_1 \p_1' + u_2 \p_2' \)
implies
\(y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'\)

Plug into \(y'' + p(t)y' + q(t)y = g(t)\):

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g\)

\(u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2' + qu_1 \p_1 + qu_2 \p_2 = g\)

\(u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g\)

\(u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g\)

\(\p_1\), \(\p_2\) are homogeneous solutions.  Thus \(\p_i'' + p \p_i' +  q\p_i = 0\).

Hence \(u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g\)

Thus we have 2 eqns to find  2 unknowns, the functions \(u_1\) and \(u_2\):
 

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right] =
\left[ \matrix{ 0\cr g }\right]$

Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}\( and  \)u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

Sect.3.6:  {\bf Guess \(y = u_1(t)e^t + u_2(t) te^t\) and solve for \(u_1\) and 
\(u_2\)}

\(y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}\).

Two unknown functions, \(u_1\) and \(u_2\), but only one equation (\(y'' - 2y' + y = e^t ln(t)\)).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in \(y''\): ~~Choose \(u_1'e^t + u_2' te^t = 0\)}

Hence \(y' = u_1e^t +  u_2(e^t + te^t)\).

and 
\(y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)\).

 
~~~~~~~~~\( = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)\).
 
~~~~~~~~~\( = u_1e^t + u_2'e^t + u_2(2e^t + te^t)\).

Solve \(y'' - 2y' + y = e^t ln(t)\)

$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

\(  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)\)

\(  u_2'    = ln(t)\)
or in other words,
\({du_2 \over dt} = ln(t)\)

Thus \(\int du_2 = \int ln(t) dt\)

\(u_2 = t ln(t) - t\).   Note only need one solution, so don't need \(+C\).

 \(y = u_1(t)e^t + [ t ln(t) - t ] te^t\)

\(u_1'e^t + u_2' te^t = 0\).  Thus \(u_1' + u_2' t = 0\).  Hence \(u_1' =  -u_2' t = -tln(t) \)

Thus \(u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  \)

Thus the general solution is 

 \(y = c_1e^{t + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}\) }

\end

\end

\end

START 100/FALL18/e1_Fall2018ansOLD.tex part 3

\nopagenumbers

Math 3600 Differential Equations Exam \#1
                               Sept 26, 2018 \hfill  SHOW ALL
WORK
~~~

5.) Solve the following 2nd order differential equations

[15]~   5a.) \(2y'' - y' + 10y= 0\) 

Guess \(y = e^{rt}\)

\(2r^2 - r + 10 = 0\).  Thus $r = {1 \pm  qrt{(-1)^2 - 4(2)(10)} \over 2(2)} = {1 \pm  qrt{(1- 80)} \over 4} = {1 \pm  qrt{- 79} \over 4}
= {1  \over 4} \pm  i{ qrt{ 79} \over 4}$

 fill

General solution: \(\underline{~~y = c_1e^{t \over 4cos({ qrt{79} \over 4}t) +  c_2e^{t \over 4}sin({ qrt{79} \over 4}t) ~~}\)}

[15]~  5b.) \(x(x'') =  (x’)^2 \)  ~~~~Hint: Let \(x' = {dx \over dt} = v\), then \(v' = {dv \over dt} ={dx \over dt}{dv \over dx}   = v{dv \over dx}\)

Also   \(x' = {dx \over dt} =v\) implies \(x'' = v' = {dv \over dt} =  {dx \over dt}{dv \over dx}  = v{dv \over dx}\)

\(x{dv \over dt} = v^2\), but this has 3 variables, so replace \({dv \over dt} =  {dx \over dt}{dv \over dx}  = v{dv \over dx}\)

\(xv{dv \over dx} = v^2\)

\({dv \over v} = {dx \over x}\)

\(ln|v| = ln|x| + C\)

\(v = Cx\)

\( {dx \over dt} = Cx\)

\( {dx \over x} = Cdt\) 

\(ln|x| = Ct + k\)

\(x = ke^{Ct}\)

Check: \( ( ke^{Ct})( kC^2e^{Ct}) = Cke^{Ct} Cke^{Ct}\)

 fill

General solution: \(\underline{~~x = ke^{Ct~~}\)}
 fill

\end

START 100/SPRING13/e1_2013ANS.txt part 5

5.)   Find the general solutions for the following three  differential equations.

[15]~  5C.)  \(3y'' (y')^2 = 1\)

Let \(v  y'\).  Then \(v' = y''\).

\(3v' (v)^2 = 1\) implies \(3{dv \over dt} (v)^2 = 1\)

 \(3{dv} (v)^2 = dt\)
 
 \(\int3 (v)^2dv = \int dt\)

\(v^3 = t + C_1\).  Thus \(v = (t + C_1)^{1 \over 3}\)

\({dy \over dt} =  (t + C_1)^{1 \over 3}\)

\({dy} =  (t + C_1)^{1 \over 3}dt\)

\(\int {dy} = \int (t + C_1)^{1 \over 3}dt\)

\(y = {3 \over 4}(t + C_1)^{4 \over 3} + C_2\)

 fill  fill 
Answer 5C.) \(\underline{~~y = {3 \over 4(t + C_1)^{4 \over 3} + C_2~~}\)}

 \end
 
 

START 100/3_8appl.tex

Mechanical Vibrations:
\(mu''(t) + \gamma u'(t) + ku(t) = F_{external, ~~m, \gamma, k \geq 0\)} 
IVP:  \(u(t_0) = u_0\), ~~, \(u'(t_0) = u_1\)

NOTE:  Positive direction points DOWN.

\(m\)=  mass,
\u
\(k\) = spring force proportionality constant,
\u
\(\gamma\) = damping force proportionality constant

\u
g = 9.8 m/sec\(^2\) or 32 ft/sec\(^2\).

Weight \(= mg\)~~~~~  \(mg - kL = 0\), ~~~~~\(F_{damping(t) = -\gamma u'(t)\)}

---

A mass of 3kg  stretches a spring 4.9 m.  If the mass is acted upon by an external force of \(40e^{-t \over 3}\) N in a medium that imparts a viscous force of 10 N when the speed of the mass is 5 m/sec.  If the mass is pulled down 1 m and set in 
 motion with an upward velocity of \({8}\) m/sec, describe the motion of the mass.
 

\(m = 3\)

 
\(|F_{damping}(t)| = |\gamma u'(t)| \Rightarrow 10  = \gamma(5)\).  Thus \(\gamma = 2\)

 
\(mg - kL = 0\)  implies \( = {mg \over L} = {3(9.8) \over 4.9} = 6\)

IVP: ~~ \(3u'' + 2u' + 6u = 40e^{-t \over 3}\), ~~\(u(0) = 1\), ~~\(u'(0) = -8\)  

IVP: ~~ \(3u'' + 2u' + 6u = 40e^{-t \over 3}\), ~~\(u(0) = 1\), ~~\(u'(0) = -8\)  

---

{\bf Step 1:}  Solve homogeneous eqn:  \(3u'' + 2u' + 6u = 0\)
 
\(u = e^{rt}\) implies \(3r^2 + 2r + 6 = 0\) implies 

$r = {-2 \pm  qrt{4 - 4(3)(6)} \over 2(3)} 
= {-2  \over 2(3)} \pm { qrt{4} qrt{1 - (3)(6)} \over 2(3)}
= -{1  \over 3} \pm i{ qrt{17} \over 3} $

Thus general homogeneous solution is

\(u(t) = c_1 e^{-{t\over 3} cos({ qrt{17} \over 3} t) + c_2e^{-{t \over 3}} sin({ qrt{17} \over 3} t)\) }

---

{\bf Step 2:}  Find a non-homogeneous soln % to \(3u'' + 2u' + 6u = 40e^{-t \over 3}\)

Section 3.6:  First find Wronskian, 

\(W( e^{-{t\over 3} cos({ qrt{17} \over 3} t), e^{-{t \over 3}} sin({ qrt{17} \over 3} t))\).}
    Too much work.  Thus,

Section 3.5:  Guess \(u(t) = Ae^{-t \over 3}\).  

Then \(u'(t) = -{A \over 3}e^{-t \over 3}\) and \(u''(t) = {A \over 9}e^{-t \over 3}\)

\(3({A \over 9}e^{-t \over 3}) + 2(-{A \over 3}e^{-t \over 3}) + 6{A}e^{-t \over 3} = 40e^{-t \over 3}\)

\({A \over 3} -{2A \over 3} + 6{A} = 40\) implies \(A - 2A + 18A = 17A = 120\).  

Thus \(A = {120 \over 17}\) and 

hence \(u(t) = {120 \over 17}e^{-t \over 3}\) is a non-homogeneous soln.

Thus general NON-homogeneous solution is
\(u(t) = e^{-{t \over 3}[c_1  cos({ qrt{17} \over 3} t) + c_2 sin({ qrt{17} \over 3} t)] +  {120 \over 17}e^{-t \over 3}\) }
\(u(t) = e^{-{t \over 3}[c_1  cos({ qrt{17} \over 3} t) + c_2 sin({ qrt{17} \over 3} t) +  {120 \over 17}]\) }

---

{\bf Step 3:}  Use initial values to find \(c_1\) and \(c_2\).

\(u(t) = e^{-{t \over 3}}[c_1  cos({ qrt{17} \over 3} t) + c_2 sin({ qrt{17} \over 3} t) +  {120 \over 17}]\)

\(u'(t) = -{1 \over 3}e^{-{t \over 3}}[c_1  cos({ qrt{17} \over 3} t) + c_2 sin({ qrt{17} \over 3} t) +  {120 \over 17} ] \)

\(u(0) = 1\):  ~~\(1 = c_1 (1) + c_2 (0) +  {120 \over 17}\) 

\(u'(0) = -8\): ~~   \(-8 =  -{1 \over 3}[c_1   +  {120 \over 17}] + c_2{ qrt{17} \over 3}\) 

  skip 10pt
~~~\(-24 =  -[  -{103 \over 17   +  {120 \over 17}] + c_2{ qrt{17}}\) }

 \(-24 =  -[ {17 \over 17}] + c_2{ qrt{17}}\) implies \(c_2{ qrt{17}} = -23\) and thus \(c_2 = -{23 \over  qrt{17}} = -{23  qrt{17} \over 17}\)

Thus solution to IVP is
\(u(t) = e^{-{t \over 3}[ -{103 \over 17}  cos({ qrt{17} \over 3} t) -{23  qrt{17} \over 17} sin({ qrt{17} \over 3} t) +  {120 \over 17}]\) }

Thus solution to IVP is
\(u(t) = e^{-{t \over 3}[ -{103 \over 17}  cos({ qrt{17} \over 3} t) -{23  qrt{17} \over 17} sin({ qrt{17} \over 3} t) +  {120 \over 17}]\) }

Simplify:
\(u(t) = e^{-{t \over 3}[ R cos\delta  cos({ qrt{17} \over 3} t) + R sin \delta sin({ qrt{17} \over 3} t) +  {120 \over 17}]\) }
\(u(t) = e^{-{t \over 3}[ R  cos({ qrt{17} \over 3} t - \delta) +  {120 \over 17}]\) }

where \(R cos\delta  = c_1 =  -{103 \over 17}\) and \(R sin \delta = c_2 = -{23  qrt{17} \over 17} \).

Thus \(R =  qrt{R^2cos^2\delta + R^2 sin^2 \delta} =  qrt{c_1^2 + c_2^2} \)

\(R  =  qrt{( -{103 \over 17)^2 + ( -{23  qrt{17} \over 17})^2}\)% = {  qrt{10609 + 8993  } \over 17} 
$= {  qrt{19602  } \over 17}
= { qrt{ 2(3)^4(11)^2} \over 17} = {99 qrt{ 2} \over 17}$}
 fill

and \({c_2 \over c_1} = {Rsin \delta \over Rcos \delta} = tan{\delta}\).  Thus \(\delta = tan^{-1}({c_2 \over c_1})\), sort of  

\(\delta = tan^{-1}({23  qrt{17} \over 103}) + \pi  im  222.64^o  im 3.8858\) radians 

Simplified answer to IVP:
\(u(t) = e^{-{t \over 3}[ {99 qrt{ 2} \over 17}  cos({ qrt{17} \over 3} t - ( tan^{-1}({23  qrt{17} \over 103}) + \pi )) +  {120 \over 17}]\) }

Approximation of solution:
\(u(t)  im e^{-{t \over 3}[ {99 qrt{ 2} \over 17}  cos({ qrt{17} \over 3} t -  {5 \pi \over 4}) +  {120 \over 17}]\) }

\end

START 100/3_8trig.tex

3.7/8 Mechanical Vibrations:
\(mu''(t) + \gamma u'(t) + ku(t) = F_{external, ~~m, \gamma, k \geq 0\)} 
 \(mg - kL = 0\), ~~~~~\(F_{damping(t) = -\gamma u'(t)\)}
\u
\(m\)=  mass,
\u
\(k\) = spring force proportionality constant,
\u
\(\gamma\) = damping force proportionality constant

\u
\(g\) = 9.8 m/sec\(^2\) or 32 ft/sec\(^2\). ~~~ Weight = \(mg\).

Electrical Vibrations:
Voltage drop across inductor +  resistor + capacitor

~~$L{dI(t) \over dt + RI(t) + {1 \over C}Q(t) = E(t), ~~L, R, C \geq 0 {\hbox{ and 
}}
I = {dQ \over dt}$~~~~~~}

\(LQ''(t) + RQ'(t) + {1 \over CQ(t) = E(t)\)}

\(L\) = inductance (henrys),
\u
\(R\) = resistance (ohms)
\u
\(C\) = capacitance (farads)
\u
\(Q(t)\) = charge at time \(t\) (coulombs)
\u
\(I(t)\) = current at time \(t\) (amperes)
\u
\(E(t)\) = impressed voltage (volts).

1 volt = 1 ohm \(\cdot\) 1 ampere = 1 coulomb / 1 farad = 1 henry \(\cdot\) 1 amperes/ 1 second

Trig background:
 
\(cos (y \mp x) = cos(x \mp y) = cos(x) cos(y) \pm sin(x)sin(y)\)

Let \(c_1 = Rcos(\delta)\),  \(c_2 = Rsin(\delta)\) in 

\(c_1cos(\omega_0t) + c_2 sin(\omega_0t)\)
\u
\hskip 0.3in = \(Rcos(\delta)cos(\omega_0t) +  Rsin(\delta) sin(\omega_0t)\)
\u
\hskip 0.3in = \(Rcos(\omega_0t - \delta)\)

Amplitude = \(R\)
\u
frequency = \(\omega_0\) (measured in radians per unit time).
\u
period = \({2\pi \over \omega_0}\)
\hfill
phase (displacement) = \(\delta\)

\(c_1 = Rcos(\delta)\),  \(c_2 = Rsin(\delta)\) implies
 
\(c_1^2 + c_2^2 = R^2cos^2(\delta)  + R^2sin^2(\delta) \)

and \({Rsin(\delta) \over Rcos(\delta)} = tan(\delta) = {c_2 \over c_1}\)

BUT easier to plot to convert Euclidean coordinates \((c_1, c_2) = (Rcos(\delta), Rsin(\delta))\)
 into polar coordinates \((R, \delta)\) = (length, angle).

3.7:  Homogeneous equation (no external force):
\(mu''(t) + \gamma u'(t) + ku(t) = 0, ~~m, \gamma, k \geq 0\)

\(r_1, r_2 = {-\gamma \pm  qrt{\gamma^2 - 4km} \over 2m}\)

---

{\bf Critical damping: \(\gamma = 2 qrt{km}\)}

\(\gamma^2 - 4km = 0\):  \(u(t) = (c_1 + c_2t)e^{r_1t} \)

~~~~Note \(r_1 = -{\gamma  \over 2m} < 0\). \hfill  {\bf Thus \(u(t) \rightarrow 0\) as \(t \rightarrow \infty\)}

---

{\bf Overdamped: \(\gamma > 2 qrt{km}\)}

\(\gamma^2 - 4km > 0\):  \(u(t) = c_1e^{r_1t} + c_2e^{r_2t}\)

~~~~Note  \(r_1, r_2 < 0\). \hfill   {\bf Thus \(u(t) \rightarrow 0\) as \(t \rightarrow \infty\)}

~~~~Example \(u(t) = 4e^{-t} - 3e^{-2t} \)%= e^{-t}(4 - 3e^{-t}) $

~~~~~~~If \( t >  0\),  \(4e^{-t} > 3e^{-2t}\)

~~~~~~~If \( t <  0\),  \(4e^{-t} < 3e^{-2t}\)

{\bf Underdamped: \(\gamma < 2 qrt{km}\)}

\(\gamma^2 - 4km < 0\):  \(u(t) = e^{-{\gamma t \over 2m}}(c_1 cos \mu t + c_2 sin \mu t)\)

\(\mu\) =  quasi frequency, \({2 \pi \over \mu}\) = quasi period

Note if \(\gamma \not= 0\), ~~~~{\bf then \(u(t) \rightarrow 0\) as \(t \rightarrow \infty\)}
 fill
Note if \(\gamma = 0\), then
 fill
---

{\bf NOTE if \(\gamma \not= 0\), then homogeneous solution goes to 0 as} \(t \rightarrow \infty\).  

Thus initial values have very little effect on the long-term behaviour of solution if  \(\gamma \not= 0\).

Note:  The larger \(\gamma\) is, the faster the  homogeneous solution goes to 0 as \(t \rightarrow \infty\). 

~

3.8:   \(F_{external} \not= 0\)
\(mu''(t) + \gamma u'(t) + ku(t) = F_{external, ~~m, \gamma, k \geq 0\)} 

General solution:  \(u(t) = c_1\phi_1 + c_2\phi_2 + \psi\)
where \(\phi_1, \phi_2\) are homogeneous solutions and \(\psi\) is a non-homogeneous solution.

 NOTE if \(\gamma \not= 0\), then homogeneous solution 

Thus  if \(\gamma \not= 0\), then \(u(t)  \rightarrow  \psi\)  as \(t \rightarrow \infty\).  

---

{\bf No damping} \((\gamma = 0)\) example:  \(u''+u=cos( t)\) 
Step 1:  Solve homogeneous \(u''+u=0\)

\(r^2 +1 = 0\) implies \(r = \pm i\)

Step 2:  Find a non-homogeneous solution.

Guess \(u(t) = \)

Plug in plus lots of work implies \(A = 0\) and \(B = {1 \over 2}\)

Thus general non-homogeneous solution:  

\(u(t) =  c_1cos(t) + c_2sin(t) +  {1 \over 2tsin(t) \)}

{\bf No damping} example  \(u''+u=cos(\omega t)\) where \(\omega\neq 1\).  

Step 1:  Solve homogeneous \(u''+u=0\)

\(r^2 +1 = 0\) implies \(r = \pm i\)

Step 2:  Find a non-homogeneous solution.

Since \(\omega\neq 1\), guess \(u(t) = \)

 fill

Trig background:

\(cos (y \mp x) = cos(x \mp y) = cos(x) cos(y) \pm sin(x)sin(y)\)

\(cos(u) + cos(v) = 2cos({u+v \over 2})cos({u-v \over 2})\)

\(cos(u) - cos(v) = -2sin({u+v \over 2})sin({u-v \over 2})\)

\(sin(u) + sin(v) = 2sin({u+v \over 2})cos({u-v \over 2})\)

\(sin(u) - sin(v) = sin(u) + sin(-v) = 2sin({u-v \over 2})cos({u+v \over 2})\)

---

Derivation:

Let \(x = ({u+v \over 2})\) and \(y = ({u-v \over 2})\) 

\( cos(u)= cos( ({u+v \over 2}) + ({u-v \over 2}))\)
\w

\( cos(v)= cos( ({u+v \over 2}) - ({u-v \over 2}))\)
\w

---

Ex:  \(u(t) = cos(t) + cos(3t) = \)

Graph:

Example {\bf  with damping: }
\(u''+\gamma u' + u=cos(\omega t)\) where \(\gamma\) is small. 

\end

Example with :  \(u''+ \gamma u' + u=0\).  I note that when gamma is postive and close to
zero, the solution dies very slowly and introduce the notion of
quasifrequency. I also notice that the oscillating behavior of the
decaying solutions slows down as we increase the friction staying near zero.

4.  Next I solve  u''+gamma u' + u=cos(omega t) and break the general
solution into its transitory and steady state solution. I write the
steady state solution in the form  amplitude times cos(omega t -delta)
where delta is the phase shift and I plot the amplitude where gamma is
fixed as small positive number.

START 100/SLIDES2/3_7worksheetANSWERS.tex

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\usepackage{amsmath}
\usepackage{graphicx}
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\begin{document}

1. The following shows the graphs of \(u = e^{-t}\), \(u = -e^{-2t}\) and \(u =  e^{-t}-e^{-2t}\).

1a.) The thin red curve corresponds to \(\underline{~~u = e^{-t}~~}\). 

1b.) The medium dotted blue curve corresponds to \(\underline{~~u = -e^{-2t}~~}\). 

1c.)  The thick light green curve  corresponds to \(\underline{~~u =  e^{-t}-e^{-2t}~~}\). 

1d.) \(\displaystyle\lim_{t \rightarrow -\infty} e^{-t}-e^{-2t} = -\infty\)

1e.) \(\displaystyle\lim_{t \rightarrow \infty} e^{-t}-e^{-2t} = 0\)

1f.)  As \({t \rightarrow -\infty}\) which term of \( e^{-t}-e^{-2t}\) dominates.  In other words, does  \( e^{-t}-e^{-2t}\) look more like  \( e^{-t}\) or  \( -e^{-2t}\)  for large negative values of \(t\)
 
\( -e^{-2t\)}

1f.)  As \({t \rightarrow \infty}\) which term of \( e^{-t}-e^{-2t}\) dominates.  In other words, does  \( e^{-t}-e^{-2t}\) look more like  \( e^{-t}\) or  \( -e^{-2t}\)  for large positive values of \(t\)
 
\( e^{-t\)}

---

2a.) Suppose \(u(t) = c_1\phi_i +c_1\phi_2\) is the general {\bf homogeneous} solution to a differential equation modeling mechanical vibration {\bf with damping}, then  \(\displaystyle\lim_{t \rightarrow \infty} u(t) = 0\)

2b.)  Will the initial values affect the long-term behaviour of the  general {\bf homogeneous} solution to a differential equation modeling mechanical vibration {\bf with damping}?
 
No.

3.)  Suppose \(u'' + 4u = tcos(2t)\) models mechanical vibration (with no damping and with external force = \(tcos(2t))\)

Normally when solving \(au'' + bu' + cu = tcos(2t)\) one would guess a non-homogenous solution to be of the form \(u = (At + B)[Ccos(2t) + Dsin(2t)]\) since  \(tcos(2t)\)
is the product of a degree 1 polynomial and \(cos(2t)\).  But that won't work in this case.

\( (At + B)[Ccos(2t) + Dsin(2t)] =(At)[Ccos(2t) + Dsin(2t)] + (B)[Ccos(2t) + Dsin(2t)]\)

3a.) \(u =  (B)[Ccos(2t) + Dsin(2t)]\) is NOT a solution to the non-homogeneous equation  \(u'' + 4u = tcos(2t)\) since it is a solution to the equation:

\(u'' + 4u = 0\)

Sidenote 1:  If you did include this in your guess, this term would cancel out since it is a homogeneous solution, and you would see that any choice of \(B\) would work IF the rest of your guess worked.  \(B = 0\) is the simplest choice. 

3b.)  \(u =  (At)[Ccos(2t) + Dsin(2t)]\) is NOT a solution to the non-homogeneous equation  \(u'' + 4u = tcos(2t)\) since it is a solution to the equation:
 
\(u'' + 4u = cos(2t)\)

Sidenote 2:  If I were solving the equation that answers 3B, I would let C and D swallow the constant A and thus guess \(u =  t[Ccos(2t) + Dsin(2t)]\) instead of \(u =  (At)[Ccos(2t) + Dsin(2t)]\)

3c.)  Thus the best guess for a non-homogeneous solution to  \(u'' + 4u = tcos(2t)\) would be

\(u = t (A t + B) [C cos(2 t) + D sin(2 t)]\) or
\(u = A_1 t^2 cos(2 t) + A_2 t^2 sin(2 t) + A_3 t cos(2 t) + A_4 t sin(2 t)\)

 
Hint:  Your answer will likely include 4 unknowns.

3d.) Plugging in \(u = t (A t + B) [C cos(2 t) + D sin(2 t)]\) would result in 4 nonlinear equations involving products such as \(AC\). Note 

$\begin{matrix}
u &=& t (A t + B) [C cos(2 t) + D sin(2 t)]\hfill \cr
&=& A C t^2 cos(2 t) + A D t^2 sin(2 t) + B C t cos(2 t) + B D t sin(2 t)  \cr
&=&  A_1 t^2 cos(2 t) + A_2 t^2 sin(2 t) + A_3 t cos(2 t) + A_4 t sin(2 t) 
\end{matrix}$

Thus we can instead plug in \(u = A_1 t^2 cos(2 t) + A_2 t^2 sin(2 t) + A_3 t cos(2 t) + A_4 t sin(2 t)\).

Solve \(u'' + 4u = tcos(2t)\)  using  that 

\([A_1 t^2 cos(2 t) + A_2 t^2 sin(2 t) + A_3 t cos(2 t) + A_4 t sin(2 t)]'' \)

~~~~~~~~~\(=  - 8 A_1 t sin(2 t)+ 2 A_2 sin(2 t) - 4 A_3 sin(2 t) + 8 A_2 t cos(2 t)  + 2 A_1 cos(2 t) + 4 A_4 cos(2 t)\)

Answer:  The LHS of  \(u'' + 4u = tcos(2t)\) is given above.  Setting LHS = RHS:

 \(- 8 A_1 t sin(2 t)+ 2 A_2 sin(2 t) - 4 A_3 sin(2 t) + 8 A_2 t cos(2 t)  + 2 A_1 cos(2 t) + 4 A_4 cos(2 t) = tcos(2t)\)

Combine like terms:

 \(- 8 A_1 t sin(2 t)+ (2 A_2  - 4 A_3) sin(2 t) + 8 A_2 t cos(2 t)  + (2 A_1  + 4 A_4) cos(2 t) \)

tcos(2t) + 0 cos(2 t) $}

Since $\{t sin(2 t), sin(2 t), 
tcos(2t), cos(2 t) \}$ is a linearly independent set, the coefficients of like terms from LHS equal the coefficients of the corresponding like terms on RHS.  Thus 

$ 8 A_1 = 0,~~ 
2 A_2  - 4 A_3 = 0,~~
 8 A_2 t = 1,~~  
2 A_1  + 4 A_4 = 0$

Thus \(A_1 = A_4 = 0\), \(A_2 = {1 \over 8}\), and \(A_3 = {1 \over 2}A_2 = ({1 \over 2})({1 \over 8}) = {1 \over 16}\)

Thus the general solution is \(u(t) = c_2 sin(2 t) + c_1 cos(2 t) + {1 \over 8}t^2 sin(2 t) + {1 \over 16} t cos(2 t)\)

\renewcommand{\arraystretch}{2.5}

4.)  The following are solutions to a second order differential equation modeling mechanical vibration.
Match these equations to their graphs.  State whether the graph corresponds to an undamped, underdamped, critically damped, or overdamped mechanical vibration.

\begin{center}
\begin{tabular}{ |c |c |c| }
 \hline
  Equation      & Graph &  damping \\ \hline
\(u = 3cos(2t + {\pi \over 6})\) & B & none \\    \hline 
\(u = 3cos(2t - {\pi \over 6})\) & A & none\\    \hline 
\(u = 3 e^{-2t}cos(2t + {\pi \over 6})\)  & F & underdamped \\    \hline 
\(u =  3e^{-2t}cos(2t - {\pi \over 6})\) & D & underdamped \\    \hline 
\(u = 3 e^{-2t}cos(5t + {\pi \over 6})\) & E & underdamped \\    \hline 
\(u = 3 e^{-2t}cos(5t - {\pi \over 6})\) & C & underdamped \\    \hline 
\(u = e^{-2t}(1 + t)\) & G & ~~critically damped ~~\\    \hline 
\(u = 100 e^{-2t} + e^{-5t}\) & I & overdamped \\    \hline 
\(u = -100 e^{-2t} - e^{-5t}\) & K & overdamped \\    \hline 
\(u = -100 e^{-2t} + e^{-5t}\) & H & overdamped \\    \hline 
\(u = 100 e^{-2t} - e^{-5t}\) & J & overdamped \\    \hline 
\end{tabular}
\end{center}

A. \includegraphics[width= 3.4in]{A \hfill
B. \includegraphics[width= 3.4in]{B} \hfill}

 
C. \includegraphics[width= 3.4in]{C \hfill
D. \includegraphics[width= 3.4in]{D} \hfill}
 
E. \includegraphics[width= 3.4in]{E \hfill
F. \includegraphics[width= 3.4in]{F} \hfill}
 
G. \includegraphics[width= 3.4in]{G \hfill
H. \includegraphics[width= 1.4in]{Rplot02} \hfill}

 
I. \includegraphics[width= 1.4in]{Rplot03 \hfill
J. \includegraphics[width= 1.4in]{Rplot04} \hfill
K. \includegraphics[width= 1.4in]{Rplot05} \hfil}

\end{document}

START 100/FALL17/quiz4Fall2017ans.tex part 1

1.  A mass weighing 5 lbs stretches a spring 8 in. The mass is acted on by an external force of 9sin(2t) lbs.  The mass is pulled down 1 foot and then set in motion with an upward velocity of 3ft/s. Assume that there is no damping. Note \(g = 32ft/s^2\).  State the initial value problem that describes the motion of this mass.

IVP:  \(\underline{~ {5 \over 32}u''  + {15 \over 2}u = 9sin(2t), ~u(0) = +2, ~u'(0) = -8 ~}\)

\(mg = 5\).  Thus \(m = {5 \over 32}\).
\hfill
\(kL = mg\).  Thus \(k(8/12) = k(2/3) =5\).   Thus \(k = {15 \over 2}\).
\(mu'' + 0u' + ku = 9sin(2t) \)~~~~~~~~~~~~~~~~~~~~~~

2.)  Given that the solution to \(y'' + y = 0\) is \(y = c_1cos(t) + c_2sin(t)\), what would be a good guess for a non-homogeneous solution to
\(y'' + y = cos(2t)\)?  Note you do not need to solve this differential equation.  You also don't need to determine the undetermined coefficients.

Acceptable guess:  \(\underline{~y = Acos(2t) + Bsin(2t) ~\)}
  skip 10pt
Best guess:  \(\underline{~y = Bcos(2t) ~\)}
Since no \(y'\) term, don't need \(sin\) term.  However, both guesses will give you the correct non-homogeneous solution, so both answers are correct since I didn't ask for the best guess.

3.)  Suppose that \(y_1(t) = t\) and \(y_2(t) = t^2\) are solutions to the differential equation, \(y'' + p(t)y' + q(t)y = 0\).  Find the general solution to   \(y'' + p(t)y' + q(t)y = {1 \over t}\)

General solution:  \(\underline{~y = c_1t + c_2t^2 -tln|t| ~}\)

\(W(t, t^2) = \left|\matrix{t  & t^2 \cr 1 & 2t} \right| = 2t^2 - t^2 = t^2\).  %~~~~~Check by Abel's theorem:  $W(t, t^2) = e^{-\int p_1(t) dt}

\({1 \over t}\left|\matrix{0 & t^2 \cr 1 & 2t} \right|  = \left|\matrix{0 & t^2 \cr {1 \over t} & 2t} \right| =-t\) \hfill 
\(u_1(t) =  \int {g(t) \over a}{W_1 \over W} = \int {-t \over t^2} dt = \int {-1 \over t} dt = - ln|t|\)

\(\left|\matrix{t  & 0 \cr 1 & {1 \over t}} \right| =1\)
\hfill 
\(u_2(t) =  \int {g(t) \over a}{W_2 \over W} = \int {1 \over t^2} dt = \int t^{-2} dt = -t^{-1}\)

Non-homog:\( -tln|t| - t^{-1}t^2 = -tln|t| - t\)

General solution: \(y = k_1t + c_2t^2 -tln|t| - t = (k_1- 1)t + c_2t^2 -tln|t|  = c_1t + c_2t^2 -tln|t|\)

FYI:  

 \(y'' + p(t)y' + q(t)y = 0\).

\(y_1(t) = t\), \(y' = 1\), \(y'' = 0\)

and \(y_2(t) = t^2\), \(y' = 2t\), \(y'' = 2\)

 \(0 + p(t) + q(t)t = 0\).

 \(2 + 2p(t)t + q(t)t^2 = 0\).

 \(0 + p(t)t + q(t)t^2 = 0\).

\(2 + tp(t) = 0\).  Thus \(p(t) = -{2 \over t}\)

\(0 +  -{2 \over t} + q(t)t = 0\).  Thus \(q(t) = {2 \over t^2}\)

 \(y'' -{2 \over t}y' +  {2 \over t^2}y = 0\).

By Abel's thm, \(W(t, t^2) = e^{\int{2 \over t} dt} = e^{2ln|t|} = e^{ln|t|^2} = t^2\) 

\end

START 100/FALL17/quiz4Fall2017formBans.tex part 3

\nopagenumbers

Quiz 4 Form B ~~~~
Oct 16, 2017

1.  A mass weighing 9 lbs stretches a spring 5 in. The mass is acted on by an external force of 7sin(3t) lbs.  The mass is pulled down  2 feet and then set in motion with an upward velocity of 8ft/s. Assume that there is no damping. Note \(g = 32ft/s^2\).  State the initial value problem that describes the motion of this mass.

IVP:  \(\underline{~ {9 \over 32}u''  + {108 \over 5}u = 7sin(3t), ~u(0) = +2, ~u'(0) = -8 ~}\)

\(mg = 9\).  Thus \(m = {9 \over 32}\).
\hfill
\(kL = mg\).  Thus \(k(5/12) = 9\).   Thus \(k = {108 \over 5}\).
\(mu'' + 0u' + ku = 7sin(3t) \)~~~~~~~~~~~~~~~~~~~~~~

\end

START 34/FALL10/e2_34_F10ANS.txt part 2

[4]~~ 3a.) Given  \(2y''  + 5y = cos(wt)\), determine the value  
\(w\) for which undamped resonance occurs:   

\(2r^2 + 5 = 0\).  Thus \(r  = i qrt{5 \over 2}\).
Hence homogeneous soln is \(y(t) = c_1 cos(t qrt{5 \over 2}) + c_2 sin(t qrt{5 \over 2})\).  Hence a potential solution for the non-homogeneous equation  \(2y''  + 5y = cos(t qrt{5 \over 2})\) would be of the form: \(t[A cos(t qrt{5 \over 2}) + B sin(t qrt{5 \over 2})]\).

Answer \(\underline{ w =  qrt{5 \over 2~~}\)}
\u\u
[3]~~ 3b.)  Briefly describe in words the long-term behaviour of a solution to \break
 \(2y''  + 5y = cos(wt)\) for this value of 
\(w\).