Loading [MathJax]/extensions/TeX/boldsymbol.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

2.2E: Exercises

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

In Exercises (2.2E.1) to (2.2E.12), find the general solution.

Exercise \PageIndex{1}

y''+5y'-6y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{2}

y''-4y'+5y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{3}

y''+8y'+7y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{4}

y''-4y'+4y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{5}

y''+2y'+10y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{6}

y''+6y'+10y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{7}

y''-8y'+16y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{8}

y''+y'=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{9}

y''-2y'+3y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{10}

y''+6y'+13y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{11}

4y''+4y'+10y=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{12}

10y''-3y'-y=0

Answer

Add texts here. Do not delete this text first.

In Exercises (2.2E.13) to (2.2E.17), solve the initial value problem.

Exercise \PageIndex{13}

y''+14y'+50y=0, \quad y(0)=2,\quad y'(0)=-17

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{14}

6y''-y'-y=0, \quad y(0)=10,\quad y'(0)=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{15}

6y''+y'-y=0, \quad y(0)=-1,\quad y'(0)=3

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{16}

4y''-4y'-3y=0, \quad y(0)=\displaystyle{13\over 12},\quad y'(0)=\displaystyle{23 \over 24}

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{17}

4y''-12y'+9y=0, \quad y(0)=3,\quad y'(0)=\displaystyle{5\over 2}

Answer

Add texts here. Do not delete this text first.

In Exercises \9(2.2E.18)\) to (2.2E.21), solve the initial value problem and graph the solution.

Exercise \PageIndex{18}

y''+7y'+12y=0, \quad y(0)=-1,\quad y'(0)=0

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{19}

y''-6y'+9y=0, \quad y(0)=0,\quad y'(0)=2

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{20}

36y''-12y'+y=0, \quad y(0)=3,\quad y'(0)=\displaystyle{5\over2}

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{21}

y''+4y'+10y=0, \quad y(0)=3,\quad y'(0)=-2

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{22}

(a) Suppose y is a solution of the constant coefficient homogeneous equation

\begin{equation}\label{eq:2.2E.1} ay''+by'+cy=0. \end{equation}

Let z(x)=y(x-x_0), where x_0 is an arbitrary real number. Show that

\begin{eqnarray*} az''+bz'+cz=0. \end{eqnarray*}

(b) Let z_1(x)=y_1(x-x_0) and z_2(x)=y_2(x-x_0), where \{y_1,y_2\} is a fundamental set of solutions of \eqref{eq:2.2E.1} (A). Show that \{z_1,z_2\} is also a fundamental set of solutions of \eqref{eq:2.2E.1}.

(c) The statement of Theorem (2.2.1) is convenient for solving an initial value problem

\begin{eqnarray*} ay''+by'+cy=0, \quad y(0)=k_0,\quad y'(0)=k_1, \end{eqnarray*}

where the initial conditions are imposed at x_0=0. However, if the initial value problem is

\begin{equation}\label{eq:2.2E.2} ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1, \end{equation}

where x_0\ne0, then determining the constants in

\begin{eqnarray*} y=c_1e^{r_1x}+c_2e^{r_2x}, \quad y=e^{r_1x}(c_1+c_2x),\mbox{ or } y=e^{\lambda x}(c_1\cos\omega x+c_2\sin\omega x) \end{eqnarray*}

(whichever is applicable) is more complicated. Use part (b) to restate Theorem (2.2.1) in a form more convenient for solving \eqref{eq:2.2E.2}.

Answer

Add texts here. Do not delete this text first.

In Exercises (2.2E.23) tp (2.2E.28), use a method suggested by Exercise (2.2E.22) to solve the initial value problem.

Exercise \PageIndex{23}

y''+3y'+2y=0, \quad y(1)=-1,\quad y'(1)=4

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{24}

y''-6y'-7y=0, \quad y(2)=-\displaystyle{1\over3},\quad y'(2)=-5

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{25}

y''-14y'+49y=0, \quad y(1)=2,\quad y'(1)=11

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{26}

9y''+6y'+y=0, \quad y(2)=2,\quad y'(2)=-\displaystyle{14\over3}

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{27}

9y''+4y=0, \quad y(\pi/4)=2,\quad y'(\pi/4)=-2

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{28}

y''+3y=0, \quad y(\pi/3)=2,\quad y'(\pi/3)=-1

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{29}

Prove: If the characteristic equation of

\begin{equation}\label{eq:2.2E.3} ay''+by'+cy=0 \end{equation}

has a repeated negative root or two roots with negative real parts, then every solution of \eqref{eq:2.2E.3} approaches zero as x\to\infty.

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{30}

Suppose the characteristic polynomial of ay''+by'+cy=0 has distinct real roots r_1 and r_2. Use a method suggested by Exercise (2.2E.22) to find a formula for the solution of

\begin{eqnarray*} ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1. \end{eqnarray*}

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{31}

Suppose the characteristic polynomial of ay''+by'+cy=0 has a repeated real root r_1. Use a method suggested by Exercise (2.2E.22) to find a formula for the solution of

\begin{eqnarray*} ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1. \end{eqnarray*}

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{32}

Suppose the characteristic polynomial of ay''+by'+cy=0 has complex conjugate roots \lambda\pm i\omega. Use a method suggested by Exercise (2.2E.22) to find a formula for the solution of

\begin{eqnarray*} ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1. \end{eqnarray*}

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{33}

Suppose the characteristic equation of

\begin{equation}\label{eq:2.2E.4} ay''+by'+cy=0 \end{equation}

has a repeated real root r_1. Temporarily, think of e^{rx} as a function of two real variables x and r.

(a) Show that

\begin{equation}\label{eq:2.2E.5} a{\partial^2\over\partial^2 x}(e^{rx})+b{\partial \over\partial x}(e^{rx}) +ce^{rx}=a(r-r_1)^2e^{rx}. \end{equation}

(b) Differentiate \eqref{eq:2.2E.5} with respect to r to obtain

\begin{equation}\label{eq:2.2E.6} a{\partial\over\partial r}\left({\partial^2\over\partial^2x}(e^{rx})\right)+b{\partial\over\partial r}\left({\partial \over\partial x}(e^{rx})\right) +c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \end{equation}

(c) Reverse the orders of the partial differentiations in the first two terms on the left side of \eqref{eq:2.2E.6} to obtain

\begin{equation}\label{eq:2.2E.7} a{\partial^2\over\partial x^2}(xe^{rx})+b{\partial\over\partial x}(xe^{rx})+c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \end{equation}

(d) Set r=r_1 in \eqref{eq:2.2E.5} and \eqref{eq:2.2E.7} to see that y_1=e^{r_1x} and y_2=xe^{r_1x} are solutions of \eqref{eq:2.2E.4}.

Answer

Add texts here. Do not delete this text first.

Exercise \PageIndex{34}

In calculus you learned that e^u, \cos u, and \sin u can be represented by the infinite series

\begin{equation}\label{eq:2.2E.8} e^u=\sum_{n=0}^\infty {u^n\over n!} =1+{u\over 1!}+{u^2\over 2!}+{u^3\over 3!}+\cdots+{u^n\over n!}+\cdots \end{equation}

\begin{equation}\label{eq:2.2E.9} \cos u=\sum_{n=0}^\infty (-1)^n{u^{2n}\over(2n)!} =1-{u^2\over2!}+{u^4\over4!}+\cdots+(-1)^n{u^{2n}\over(2n)!} +\cdots, \end{equation}

and

\begin{equation}\label{eq:2.2E.10} \sin u=\sum_{n=0}^\infty (-1)^n{u^{2n+1}\over(2n+1)!} =u-{u^3\over3!}+{u^5\over5!}+\cdots+(-1)^n {u^{2n+1}\over(2n+1)!} +\cdots \end{equation}

for all real values of u. Even though you have previously considered \eqref{eq:2.2E.8} only for real values of u, we can set u=i\theta, where \theta is real, to obtain

\begin{equation}\label{eq:2.2E.11} e^{i\theta}=\sum_{n=0}^\infty {(i\theta)^n\over n!}. \end{equation}

Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in \eqref{eq:2.2E.11} converges for all real \theta.

(a) Recalling that i^2=-1, write enough terms of the sequence \{i^n\} to convince yourself that the sequence is repetitive:

\begin{eqnarray*} 1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,\cdots. \end{eqnarray*}

Use this to group the terms in \eqref{eq:2.2E.11} as

\begin{eqnarray*} e^{i\theta}&=&\left(1-{\theta^2\over2}+{\theta^4\over4}+\cdots\right) +i\left(\theta-{\theta^3\over3!}+{\theta^5\over5!}+\cdots\right)\\ &=&\sum_{n=0}^\infty (-1)^n{\theta^{2n}\over(2n)!} +i\sum_{n=0}^\infty (-1)^n{\theta^{2n+1}\over(2n+1)!}. \end{eqnarray*}

By comparing this result with \eqref{eq:2.2E.9} and \eqref{eq:2.2E.10}, conclude that

\begin{equation}\label{eq:2.2E.12} e^{i\theta}=\cos\theta+i\sin\theta. \end{equation}

This is Euler's Identity.

(b) Starting from

\begin{eqnarray*} e^{i\theta_1}e^{i\theta_2}=(\cos\theta_1+i\sin\theta_1), (\cos\theta_2+i\sin\theta_2), \end{eqnarray*}

collect the real part (the terms not multiplied by i) and the imaginary part (the terms multiplied by i) on the right, and use the trigonometric identities

\begin{eqnarray*} \cos(\theta_1+\theta_2)&=&\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\\ \sin(\theta_1+\theta_2)&=&\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2 \end{eqnarray*}

to verify that

\begin{eqnarray*} e^{i(\theta_1+\theta_2)}=e^{i\theta_1}e^{i\theta_2}, \end{eqnarray*}

as you would expect from the use of the exponential notation e^{i\theta}.

(c) If \alpha and \beta are real numbers, define

\begin{equation}\label{eq:2.2E.13} e^{\alpha+i\beta}=e^\alpha e^{i\beta}=e^\alpha(\cos\beta+i\sin\beta). \end{equation}

Show that if z_1=\alpha_1+i\beta_1 and z_2=\alpha_2+i\beta_2 then

\begin{eqnarray*} e^{z_1+z_2}=e^{z_1}e^{z_2}. \end{eqnarray*}

(d) Let a, b, and c be real numbers, with a\ne0. Let z=u+iv where u and v are real-valued functions of x. Then we say that z is a solution of

\begin{equation}\label{eq:2.2E.14} ay''+by'+cy=0 \end{equation}

if u and v are both solutions of \eqref{eq:2.2E.14}. Use Theorem (2.2.1) (c) to verify that if the characteristic equation of \eqref{eq:2.2E.14} has complex conjugate roots \lambda\pm i\omega then z_1=e^{(\lambda+i\omega)x} and z_2=e^{(\lambda-i\omega)x} are both solutions of \eqref{eq:2.2E.14}.

Answer

Add texts here. Do not delete this text first.


This page titled 2.2E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.

Support Center

How can we help?