1.3 - Classification of Differential Equations
( \newcommand{\kernel}{\mathrm{null}\,}\)
Categorizing differential equations by different types will inform our solution techniques. The most important features are
- The number of dependent/independent variables,
- The highest-order derivative present,
- Whether the equation is linear.
For the first feature, we have the following definitions.
A differential equation with a single independent variable and a single dependent variable.
dydt=ay+b
A differential equation with several independent variables and a single dependent variable.
∂xy∂x=∂xy∂y
A collection of differential equations with one independent variable and several dependent variables.
dx1dt=3x1−2x2,dx2dt=−2x1+x2.
Note that x1(t) and x2(t) are functions of t.
Toward the second feature, we have the important definition below.
The order of a differential equation is the order of the highest derivative present in the equation.
- dvdt=g−γmv is a 1-st order (ordinary) differential equation.
- md2xdt2+kx=0 is a 2-nd order (ordinary) differential equation.
- ∂u∂t=α∂2u∂x2 is a 2-nd order (partial) differential equation called the Heat Equation.
- y(n)=f(t,y,…,y(n−1)) is the general expression of an n-th order (ordinary) differential equation.
The last feature is the biggest factor toward our solution techniques, and borrows a concept from Linear Algebra.
An ordinary differential equation is linear if it can be written as a linear combination involving y,y′,…,y(n). A general linear n-th order differential equation has the form
an(t)y(n)+an−1(t)y(n−1)+⋯+a1(t)y′+a0(t)y=g(t),
for some coefficient functions ai(t). If an equation cannot be written in this way, it is non-linear.
Below are examples of linear equations:
- 2d2xdt2+2tdxdt+5=10cos(t).
- t3y′+3t2y=4.
- dvdt=g−γmv.
Below are examples of non-linear equations:
- dydt=0.5(1−y3)y is non-linear because of the y2 term after distributing.
- sin(dydx)=xy is non-linear (only) because of the sin(dydx) term.
Determine the linearity of the following differential equations.
1) ty″−t3y′−3y=sin(t)
2) 2y″−3y′−3y2=0
- Answer
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Both equations are linear since only y and its derivatives y′ and y″ are present, and not functions of them.
Q1) For which value of r is y(t)=ert a solution of the linear 1-st order differential equation 2y′−6y=0?
Q2) Is y(t)=Ce3t also a solution?
- Hint
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A function y(t) is a solution if the differential equation is true after 'plugging in' y(t).
- Answer Q1
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In order to check if y is a solution, we also need its derivative y′=ddty=ddtert=rert. In order for y to be a solution, we need the following to be true:
2y′−6y=0⟺2(rert)−6(ert)=0,⟺(2r−6)ert=0,⟺2r−6=0.(Sinceertis never zero).
Thus, y(t)=ert is a solution to the differential equation only if r=3.
- Answer Q2
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For y(t)=Ce3t we have y′=ddtCe3t=Cddte3t=3Cert. We see that
2y′−6y=0⟺2(3Cert)−6(Cert)=0,⟺(6−6)Cert=0,
which is true regardless of the value of C.
The answer to Q2 in the previous example shows an interesting feature of linear differential equations: Cy is a solution whenever y is. If we want to further narrow down our solutions, we need more information.
An initial value problem is an n-th order differential equation together with initial conditions on its solutions and their lower-order derivatives. Initial value problems may have no solutions, a unique solution, or infinitely-many solutions.
Below are examples of IVPs for 1-st order differential equations.
- dvdt=g−γmv;v(0)=60.
- 2y′−6y=0;y(0)=4.
Below are examples of IVPs for 2-nd order differential equations.
- d2xdt2+4x=0;x(0)=1,x′(0)=0.
- y″+2y′+5y=cos(t);y(0)=0,y′(0)=2.
We will make this more precise in future chapters, but generically an n-th order IVP with n-many initial conditions (one for each lower-order derivative) admits a unique solution.
Determine the unique solution of the IVP 2y′−6y=0;y(0)=4.
- Answer
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From exercise 2 above, we know y=Ce3t is a solution to 2y′−6y=0.
Plug in initial value to find C :
If y(0)=4, then 4=Ce3(0) implies C=4.
Thus by existence and uniqueness theorem, y=4e3t is the unique solution to IVP: 2y′+6y=0,y(0)=4.
Suppose salty water with a salt concentration of 3[gL] enters and leaves a tank at a rate of 2[Lmin]. If the tank contains 4[L] of water and initially contains 5[g] of salt, find a formula for the amount of salt in the tank after t minutes.
Solution
Let Q(t) be the amount of salt in the tank measured in grams [g]. This amount is probably changing with time, so let's try to construct an expression for dQdt. Also since we are tracking something inside the tank, the expression for its rate of change is likely
dQdt=‘‘Rate of salt flowing in"−‘‘Rate of salt flowing out"
Both sides of this expression must have units [gmin]. The first two pieces of data are related to something "flowing in", and their product has the right units:
‘‘Rate of salt flowing in"=3[gL]⋅2[Lmin]=6[gmin].
The only other "flowing out" data is that the water flow rate is the same. We need a [gL] term, which has the type of units of a concentration. We can use Q(t) and the volume of the tank:
‘‘Rate of salt flowing out"=Q(t)4[gL]⋅2[Lmin]=4Q[gmin].
Putting it all together, and including the initial condition Q(0)=5[g], we have
dQdt=6−Q2;Q(0)=5.
This resulting equation is a linear 1-st order IVP. We will learn techniques to solve this differential equation in the following chapter.
(Bump the following to S2.3?)
Using the easier 2.2 (matter of opinion):
dQdt=6−Q2⇒2dQ=(12−Q)dt⇒2dQ12−Q=dt∫2dQ12−Q=∫dt⇒−2ln|12−Q|=t+Cln|12−Q|=−t2+C⇒|12−Q|=e−t2+C=e−t2eC12−Q=±eCe−t2=Ce−t2⇒Q=Ce−t2+12Q(0)=5:5=Ce0+12=C+12. Thus C=−7 Thus Q(t)=−7e−t2+12
We then see that the long term behaviour of the solution lim
Thus the amount of salt in tank approaches 12 \mathrm{~g} as t \rightarrow \infty. Hence the concentration of salt approaches:
\notag
12 \mathrm{~g} / 4 \mathrm{~L}=3 \mathrm{~g} / \mathrm{L} \text { as } t \rightarrow \infty \text {. }