2.3E: Exercises

In Exercises $(2.3E.1)$ to $(2.3E.6)$, find a particular solution by the method used in Example $(2.3.2)$. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

Exercise $\PageIndex{1}$

$y''+5y'-6y=22+18x-18x^2$

Answer

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Exercise $\PageIndex{2}$

$y''-4y'+5y=1+5x$

Answer

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Exercise $\PageIndex{3}$

$y''+8y'+7y=-8-x+24x^2+7x^3$

Answer

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Exercise $\PageIndex{4}$

$y''-4y'+4y=2+8x-4x^2$

Answer

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Exercise $\PageIndex{5}$

$y''+2y'+10y=4+26x+6x^2+10x^3, \quad y(0)=2, \quad y'(0)=9$

Answer

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Exercise $\PageIndex{6}$

$y''+6y'+10y=22+20x, \quad y(0)=2,\; y'(0)=-2$

Answer

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Exercise $\PageIndex{7}$

Show that the method used in Example $(2.3.2)$ won't yield a particular solution of

$$\begin{equation}\label{eq:2.3E.1} y''+y'=1+2x+x^2; \end{equation}$$

that is, $\eqref{eq:2.3E.1}$ doesn't have a particular solution of the form $y_p=A+Bx+Cx^2$, where $A$, $B$, and $C$ are constants.

Answer

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In Exercises $(2.3E.8)$ to $(2.3E.13)$, find a particular solution by the method used in Example $(2.3.3)$.

Exercise $\PageIndex{8}$

$x^2y''+7xy'+8y=\displaystyle{6\over x}$

Answer

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Exercise $\PageIndex{9}$

$x^2y''-7xy'+7y=13x^{1/2}$

Answer

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Exercise $\PageIndex{10}$

$x^2y''-xy'+y=2x^3$

Answer

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Exercise $\PageIndex{11}$

$x^2y''+5xy'+4y=\displaystyle{1\over x^3}$

Answer

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Exercise $\PageIndex{12}$

$x^2y''+xy'+y=10x^{1/3}$

Answer

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Exercise $\PageIndex{13}$

$x^2y''-3xy'+13y=2x^4$

Answer

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Exercise $\PageIndex{14}$

Show that the method suggested for finding a particular solution in Exercises $(2.3E.8)$ to $(2.3E.13)$ won't yield a particular solution of

$$\begin{equation}\label{eq:2.3E.2} x^2y''+3xy'-3y={1\over x^3}; \end{equation}$$

that is, $\eqref{eq:2.3E.2}$ doesn't have a particular solution of the form $y_p=A/x^3$.

Answer

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Exercise $\PageIndex{15}$

Prove: If $a$, $b$, $c$, $\alpha$, and $M$ are constants and $M\ne0$ then

$$\begin{eqnarray*} ax^2y''+bxy'+cy=M x^\alpha \end{eqnarray*}$$

has a particular solution $y_p=Ax^\alpha$ ($A=$ constant) if and only if $a\alpha(\alpha-1)+b\alpha+c\ne0$.

Answer

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If $a$, $b$, $c$, and $\alpha$ are constants, then

$$\begin{eqnarray*} a(e^{\alpha x})''+b(e^{\alpha x})'+ce^{\alpha x}=(a\alpha^2+b\alpha+c)e^{\alpha x}. \end{eqnarray*}$$

Use this in Exercises $(2.3E.16)$ to $(2.3E.21)$ to find a particular solution. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

Exercise $\PageIndex{16}$

$y''+5y'-6y=6e^{3x}$

Answer

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Exercise $\PageIndex{17}$

$y''-4y'+5y=e^{2x}$

Answer

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Exercise $\PageIndex{18}$

$y''+8y'+7y=10e^{-2x}, \quad y(0)=-2,\; y'(0)=10$

Answer

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Exercise $\PageIndex{19}$

$y''-4y'+4y=e^{x}, \quad y(0)=2,\quad y'(0)=0$

Answer

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Exercise $\PageIndex{20}$

$y''+2y'+10y=e^{x/2}$

Answer

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Exercise $\PageIndex{21}$

$y''+6y'+10y=e^{-3x}$

Answer

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Exercise $\PageIndex{22}$

Show that the method suggested for finding a particular solution in Exercises $(2.3E.16)$ to $(2.3E.21)$ won't yield a particular solution of

$$\begin{equation}\label{eq:2.3E.3} y''-7y'+12y=5e^{4x}; \end{equation}$$

that is, $\eqref{eq:2.3E.3}$ doesn't have a particular solution of the form $y_p=Ae^{4x}$.

Answer

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Exercise $\PageIndex{23}$

Prove: If $\alpha$ and $M$ are constants and $M\ne0$ then constant coefficient equation

$$\begin{eqnarray*} ay''+by'+cy=M e^{\alpha x} \end{eqnarray*}$$

has a particular solution $y_p=Ae^{\alpha x}$ ($A=$ constant) if and only if $e^{\alpha x}$ isn't a solution of the complementary equation.

Answer

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If $\omega$ is a constant, differentiating a linear combination of $\cos\omega x$ and $\sin\omega x$ with respect to $x$ yields another linear combination of $\cos\omega x$ and $\sin\omega x$. In Exercises $(2.3E.24)$ to $(2.3E.29)$ use this to find a particular solution of the equation. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

Exercise $\PageIndex{24}$

$y''-8y'+16y=23\cos x-7\sin x$

Answer

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Exercise $\PageIndex{25}$

$y''+y'=-8\cos2x+6\sin2x$

Answer

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Exercise $\PageIndex{26}$

$y''-2y'+3y=-6\cos3x+6\sin3x$

Answer

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Exercise $\PageIndex{27}$

$y''+6y'+13y=18\cos x+6\sin x$

Answer

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Exercise $\PageIndex{28}$

$y''+7y'+12y=-2\cos2x+36\sin2x, \quad y(0)=-3,\quad y'(0)=3$

Answer

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Exercise $\PageIndex{29}$

$y''-6y'+9y=18\cos3x+18\sin3x, \quad y(0)=2,\quad y'(0)=2$

Answer

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Exercise $\PageIndex{30}$

Find the general solution of

$$\begin{eqnarray*} y''+\omega_0^2y =M\cos\omega x+N\sin\omega x, \end{eqnarray*}$$

where $M$ and $N$ are constants and $\omega$ and $\omega_0$ are distinct positive numbers.

Answer

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Exercise $\PageIndex{31}$

Show that the method suggested for finding a particular solution in Exercises $(2.3E.24)$ to $(2.3E.29)$ won't yield a particular solution of

$$\begin{equation}\label{eq:2.3E.4} y''+y=\cos x+\sin x; \end{equation}$$

that is, $\eqref{eq:2.3E.4}$ does not have a particular solution of the form $y_p=A\cos x+B\sin x$.

Answer

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Exercise $\PageIndex{32}$

Prove: If $M$, $N$ are constants (not both zero) and $\omega>0$, the constant coefficient equation

$$\begin{equation}\label{eq:2.3E.5} ay''+by'+cy=M\cos\omega x+N\sin\omega x \end{equation}$$

has a particular solution that's a linear combination of $\cos\omega x$ and $\sin\omega x$ if and only if the left side of $\eqref{eq:2.3E.5}$ is not of the form $a(y''+\omega^2y)$, so that $\cos\omega x$ and $\sin\omega x$ are solutions of the complementary equation.

Answer

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In Exercises $(2.3E.33)$ to $2.3E.38)$, refer to the cited exercises and use the principal of superposition to find a particular solution. Then find the general solution.

Exercise $\PageIndex{33}$

$y''+5y'-6y=22+18x-18x^2+6e^{3x}$

(See Exercises $(2.3E.1)$ and $(2.3E.16)$.)

Answer

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Exercise $\PageIndex{34}$

$y''-4y'+5y=1+5x+e^{2x}$

(See Exercises $2.3E.2)$ and $(2.3E.17)$.)

Answer

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Exercise $\PageIndex{35}$

$y''+8y'+7y=-8-x+24x^2+7x^3+10e^{-2x}$

(See Exercises $(2.3E.3)$ and $(2.3E.18)$.)

Answer

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Exercise $\PageIndex{36}$

$y''-4y'+4y=2+8x-4x^2+e^{x}$

(See Exercises $(2.3E.4)$ and $(2.3E.19)$.)

Answer

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Exercise $\PageIndex{37}$

$y''+2y'+10y=4+26x+6x^2+10x^3+e^{x/2}$

(See Exercises $(2.3E.5)$ and $(2.3E.20)$.)

Answer

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Exercise $\PageIndex{38}$

$y''+6y'+10y=22+20x+e^{-3x}$

(See Exercises $(2.3E.6)$ and $(2.3E.21)$.)

Answer

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Exercise $\PageIndex{39}$

Prove: If $y_{p_1}$ is a particular solution of

$$\begin{eqnarray*} P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x) \end{eqnarray*}$$

on $(a,b)$ and $y_{p_2}$ is a particular solution of

$$\begin{eqnarray*} P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x) \end{eqnarray*}$$

on $(a,b)$, then $y_p=y_{p_1}+y_{p_2}$ is a solution of

$$\begin{eqnarray*} P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x) \end{eqnarray*}$$

on $(a,b)$.

Answer

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Exercise $\PageIndex{40}$

Suppose $p$, $q$, and $f$ are continuous on $(a,b)$. Let $y_1$, $y_2$, and $y_p$ be twice differentiable on $(a,b)$, such that $y=c_1y_1+c_2y_2+y_p$ is a solution of

$$\begin{eqnarray*} y''+p(x)y'+q(x)y=f \end{eqnarray*}$$

on $(a,b)$ for every choice of the constants $c_1,c_2$. Show that $y_1$ and $y_2$ are solutions of the complementary equation on $(a,b)$.

Answer

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