
2.3E: Exercises

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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 $$\PageIndex{1}$$

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

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Exercise $$\PageIndex{2}$$

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

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Exercise $$\PageIndex{3}$$

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

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Exercise $$\PageIndex{4}$$

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

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Exercise $$\PageIndex{5}$$

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

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Exercise $$\PageIndex{6}$$

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

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Exercise $$\PageIndex{7}$$

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

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

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.

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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 $$\PageIndex{8}$$

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

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Exercise $$\PageIndex{9}$$

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

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Exercise $$\PageIndex{10}$$

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

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Exercise $$\PageIndex{11}$$

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

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Exercise $$\PageIndex{12}$$

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

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Exercise $$\PageIndex{13}$$

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

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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

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

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

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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 $$a\alpha(\alpha-1)+b\alpha+c\ne0$$.

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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 $$\PageIndex{16}$$

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

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Exercise $$\PageIndex{17}$$

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

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Exercise $$\PageIndex{18}$$

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

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Exercise $$\PageIndex{19}$$

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

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Exercise $$\PageIndex{20}$$

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

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Exercise $$\PageIndex{21}$$

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

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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

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

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

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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.

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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 $$\PageIndex{24}$$

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

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Exercise $$\PageIndex{25}$$

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

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Exercise $$\PageIndex{26}$$

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

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Exercise $$\PageIndex{27}$$

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

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Exercise $$\PageIndex{28}$$

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

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Exercise $$\PageIndex{29}$$

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

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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.

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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

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

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

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Exercise $$\PageIndex{32}$$

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

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

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.

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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 $$(2.3E.16)$$.)

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Exercise $$\PageIndex{34}$$

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

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

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Exercise $$\PageIndex{35}$$

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

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

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Exercise $$\PageIndex{36}$$

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

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

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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 $$(2.3E.20)$$.)

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Exercise $$\PageIndex{38}$$

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

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

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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 $$(a,b)$$.

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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 $$(a,b)$$.