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# 2.2E: Exercises

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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 $$\PageIndex{5}$$

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

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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 $$\PageIndex{11}$$

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

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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 $$\PageIndex{13}$$

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

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## Exercise $$\PageIndex{14}$$

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

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## Exercise $$\PageIndex{15}$$

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

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## Exercise $$\PageIndex{16}$$

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

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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 $$\PageIndex{18}$$

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

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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*}
\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}
\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}.

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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 $$\PageIndex{23}$$

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

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## Exercise $$\PageIndex{24}$$

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

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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 $$\PageIndex{28}$$

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

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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 $$x\to\infty$$.

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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*}
\end{eqnarray*}

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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*}
\end{eqnarray*}

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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*}
\end{eqnarray*}

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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}.

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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}.