2.2E: Exercises

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

Exercise $\PageIndex{1}$

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

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Exercise $\PageIndex{2}$

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

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Exercise $\PageIndex{3}$

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

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Exercise $\PageIndex{4}$

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

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Exercise $\PageIndex{5}$

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

Answer

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Exercise $\PageIndex{6}$

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

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Exercise $\PageIndex{7}$

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

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Exercise $\PageIndex{8}$

$y''+y'=0$

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Exercise $\PageIndex{9}$

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

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Exercise $\PageIndex{10}$

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

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Exercise $\PageIndex{11}$

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

Answer

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Exercise $\PageIndex{12}$

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

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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

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Exercise $\PageIndex{14}$

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

Answer

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Exercise $\PageIndex{15}$

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

Answer

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Exercise $\PageIndex{16}$

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

Answer

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Exercise $\PageIndex{17}$

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

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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

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Exercise $\PageIndex{19}$

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

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Exercise $\PageIndex{20}$

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

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Exercise $\PageIndex{21}$

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

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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

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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

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Exercise $\PageIndex{24}$

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

Answer

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Exercise $\PageIndex{25}$

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

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Exercise $\PageIndex{26}$

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

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Exercise $\PageIndex{27}$

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

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Exercise $\PageIndex{28}$

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

Answer

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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

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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

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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

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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

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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

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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

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