2.4E: Exercises

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In Exercises $(2.4E.1)$ to $(2.4E.14)$, find a particular solution.

Exercise $\PageIndex{1}$

$y''-3y'+2y=e^{3x}(1+x)$

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Exercise $\PageIndex{2}$

$y''-6y'+5y=e^{-3x}(35-8x)$

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Exercise $\PageIndex{3}$

$y''-2y'-3y=e^x(-8+3x)$

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Exercise $\PageIndex{4}$

$y''+2y'+y=e^{2x}(-7-15x+9x^2)$

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Exercise $\PageIndex{5}$

$y''+4y=e^{-x}(7-4x+5x^2)$

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Exercise $\PageIndex{6}$

$y''-y'-2y=e^x(9+2x-4x^2)$

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Exercise $\PageIndex{7}$

$y''-4y'-5y=-6xe^{-x}$

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Exercise $\PageIndex{8}$

$y''-3y'+2y=e^x(3-4x)$

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Exercise $\PageIndex{9}$

$y''+y'-12y=e^{3x}(-6+7x)$

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Exercise $\PageIndex{10}$

$2y''-3y'-2y=e^{2x}(-6+10x)$

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Exercise $\PageIndex{11}$

$y''+2y'+y=e^{-x}(2+3x)$

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Exercise $\PageIndex{12}$

$y''-2y'+y=e^x(1-6x)$

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Exercise $\PageIndex{13}$

$y''-4y'+4y=e^{2x}(1-3x+6x^2)$

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Exercise $\PageIndex{14}$

$9y''+6y'+y=e^{-x/3}(2-4x+4x^2)$

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In Exercises $(2.4E.15)$ to $(2.4E.19)$, find the general solution.

Exercise $\PageIndex{15}$

$y''-3y'+2y=e^{3x}(1+x)$

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Exercise $\PageIndex{16}$

$y''-6y'+8y=e^x(11-6x)$

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Exercise $\PageIndex{17}$

$ y''+6y'+9y=e^{2x}(3-5x)$

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Exercise $\PageIndex{18}$

$y''+2y'-3y=-16xe^x$

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Exercise $\PageIndex{19}$

$y''-2y'+y=e^x(2-12x)$

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In Exercises $(2.4E.20)$ to $(2.4E.23)$, solve the initial value problem and plot the solution.

Exercise $\PageIndex{20}$

$y''-4y'-5y=9e^{2x}(1+x), \quad y(0)=0,\quad y'(0)=-10$

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Exercise $\PageIndex{21}$

$y''+3y'-4y=e^{2x}(7+6x), \quad y(0)=2,\quad y'(0)=8$

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Exercise $\PageIndex{22}$

$y''+4y'+3y=-e^{-x}(2+8x), \quad y(0)=1,\quad y'(0)=2$

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Exercise $\PageIndex{23}$

$y''-3y'-10y=7e^{-2x}, \quad y(0)=1,\quad y'(0)=-17$

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In Exercises $(2.4E.24)$ to $(2.4E.29)$, use the principle of superposition to find a particular solution.

Exercise $\PageIndex{24}$

$y''+y'+y=xe^x+e^{-x}(1+2x)$

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Exercise $\PageIndex{25}$

$y''-7y'+12y=-e^x(17-42x)-e^{3x}$

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Exercise $\PageIndex{26}$

$y''-8y'+16y=6xe^{4x}+2+16x+16x^2$

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Exercise $\PageIndex{27}$

$y''-3y'+2y=-e^{2x}(3+4x)-e^x$

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Exercise $\PageIndex{28}$

$y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)$

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Exercise $\PageIndex{29}$

$y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)$

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Exercise $\PageIndex{30}$

(a) Prove that $y$ is a solution of the constant coefficient equation

\begin{equation}\label{eq:2.4E.1}
ay''+by'+cy=e^{\alpha x}G(x)
\end{equation}

if and only if $y=ue^{\alpha x}$, where $u$ satisfies

\begin{equation}\label{eq:2.4E.2}
au''+p'(\alpha)u'+p(\alpha)u=G(x)
\end{equation}

and $p(r)=ar^2+br+c$ is the characteristic polynomial of the complementary equation

\begin{eqnarray*}
ay''+by'+cy=0.
\end{eqnarray*}

For the rest of this exercise, let $G$ be a polynomial. Give the requested proofs for the case where

\begin{eqnarray*}
G(x)=g_0+g_1x+g_2x^2+g_3x^3.
\end{eqnarray*}

(b) Prove that if $e^{\alpha x}$ isn't a solution of the complementary equation then \eqref{eq:2.4E.2} has a particular solution of the form $u_p=A(x)$, where $A$ is a polynomial of the same degree as $G$, as in Example $(2.4.4)$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $y_p=e^{\alpha x}A(x)$.

(c) Show that if $e^{\alpha x}$ is a solution of the complementary equation and $xe^{\alpha x}$ isn't, then \eqref{eq:2.4E.2} has a particular solution of the form $u_p=xA(x)$, where $A$ is a polynomial of the same degree as $G$, as in Example $(2.4.5)$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $y_p=xe^{\alpha x}A(x)$.

(d) Show that if $e^{\alpha x}$ and $xe^{\alpha x}$ are both solutions of the complementary equation then \eqref{eq:2.4E.2} has a particular solution of the form $u_p=x^2A(x)$, where $A$ is a polynomial of the same degree as $G$, and $x^2A(x)$ can be obtained by integrating $G/a$ twice, taking the constants of integration to be zero, as in Example $(2.4.6)$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $y_p=x^2e^{\alpha x}A(x)$.

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Exercises $(2.4E.31)$ to $(2.4E.36)$ treat the equations considered in Examples $(2.4.1)$ to $(2.4.6)$. Substitute the suggested form of $y_p$ into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in $y_p$. Then solve for the coefficients to obtain $y_p$. Compare the work you've done with the work required to obtain the same results in Examples $(2.4.1)$ to $(2.4.6)$.

Exercise $\PageIndex{31}$

Compare with Example $(2.4.1)$:

\begin{eqnarray*}
y''-7y'+12y=4e^{2x};\quad y_p=Ae^{2x}
\end{eqnarray*}

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{32}$

Compare with Example $(2.4.2)$:

\begin{eqnarray*}
y''-7y'+12y=5e^{4x};\quad y_p=Axe^{4x}
\end{eqnarray*}

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{33}$

Compare with Example $(2.4.3)$:

\begin{eqnarray*}
y''-8y'+16y=2e^{4x};\quad y_p=Ax^2e^{4x}
\end{eqnarray*}

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{34}$

Compare with Example $(2.4.4)$:

\begin{eqnarray*}
y''-3y'+2y=e^{3x}(-1+2x+x^2),\quad y_p=e^{3x}(A+Bx+Cx^2)
\end{eqnarray*}

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{35}$

Compare with Example $(2.4.5)$:

\begin{eqnarray*}
y''-4y'+3y=e^{3x}(6+8x+12x^2),\quad y_p=e^{3x}(Ax+Bx^2+Cx^3)
\end{eqnarray*}

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{36}$

Compare with Example $(2.4.6)$:

\begin{eqnarray*}
4y''+4y'+y=e^{-x/2}(-8+48x+144x^2),\quad y_p=e^{-x/2}(Ax^2+Bx^3+Cx^4)
\end{eqnarray*}

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{37}$

Write $y=ue^{\alpha x}$ to find the general solution.

(a) $y''+2y'+y=\displaystyle{e^{-x}\over\sqrt x}$

(b) $y''+6y'+9y=e^{-3x}\ln x$

(d) $4y''+4y'+y=\displaystyle{4e^{-x/2}\left({1\over x}+x\right)}$

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{38}$

Suppose $\alpha\ne0$ and $k$ is a positive integer. In most calculus books integrals like $\int x^k e^{\alpha x}\,dx$ are evaluated by integrating by parts $k$ times. This exercise presents another method. Let

\begin{eqnarray*}
y=\int e^{\alpha x}P(x)\,dx
\end{eqnarray*}

with

\begin{eqnarray*}
P(x)=p_0+p_1x+\cdots+p_kx^k, \mbox{\quad (where $p_k\ne0$)}.
\end{eqnarray*}

(a) Show that $y=e^{\alpha x}u$, where

\begin{equation}\label{eq:2.4E.3}
u'+\alpha u=P(x).
\end{equation}

(b) Show that \eqref{eq:2.4E.3} has a particular solution of the form

\begin{eqnarray*}
u_p=A_0+A_1x+\cdots+A_kx^k,
\end{eqnarray*}

where $A_k$, $A_{k-1}$, $\dots$, $A_0$ can be computed successively by equating coefficients of $x^k,x^{k-1}, \dots,1$ on both sides of the equation

\begin{eqnarray*}
u_p'+\alpha u_p=P(x).
\end{eqnarray*}

\begin{eqnarray*}
\int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c,
\end{eqnarray*}

where $c$ is a constant of integration.

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{39}$

Use the method of Exercise $(2.4E.38)$ to evaluate the integral.

(a) $ \int e^x(4+x)\,dx$

(b) $\int e^{-x}(-1+x^2)\,dx$

(d) $\int e^x(1+x)^2\,dx$

(e) $\int e^{3x}(-14+30x+27x^2)\,dx$

(f) $\int e^{-x}(1+6x^2-14x^3+3x^4)\,dx$

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{40}$

Use the method suggested in Exercise $(2.4E.38)$ to evaluate $\int x^ke^{\alpha x}\,dx$, where $k$ is an arbitrary positive integer and $\alpha\ne0$.

Answer: Add texts here. Do not delete this text first.

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