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\begin{document}
\title{Solutions, 316-IX}
\maketitle

\section{Problem 4.8.4}
Find a particular solution to the DE
\[ x''(t) - x(t) = 3e^{-2t} \ . \]
{\bf Solution:}\\
The characteristic equation is $r^2 - 1 = 0$ with roots $r = \pm 1$. Since
$-2$ is not a root, try
\begin{eqnarray*}
 x_p(t) & = & A e^{-2t} \\
x_p^{\prime}(t) & = & -2 A e^{-2t} \\
x_p^{{\prime}{\prime}}(t) & = & 4 A e^{-2t}
\end{eqnarray*}
so that
\[ x'' - x = (4 A - A)e^{-2t} = 3e^{-2t} \]
i.e.
\[ A = 1 \rightarrow x_p(t) = e^{-2t} \ . \]
\newpage
\section{Problem 4.8.35}
Determine the form of a particular solution to the DE
\[ x'' - x' -2x = e^t\cos{t} -t^2 + t +1 \ . \]
{\bf Solution:} \\
The characteristic equation is $r^2 -r- 2 = 0$ with roots $r = 1, -2$. Since
$0$ and $1+i$ are not roots, try
\begin{eqnarray*}
 x_p(t) & = & e^{t}(A \cos{t}+B\sin{t}) +Ct^2 + Dt + E \\
x_p^{\prime}(t) & = & e^{t}\left[(A+B)\cos{t} + (-A+B)\sin{t}\right]
+ 2Ct + D \\
x_p^{\prime\prime}(t) & = & e^{t}\left[(2B)\cos{t} + (-2A)\sin{t}\right]
+ 2C
\end{eqnarray*}
so that
\[ x'' -x' -2 x= e^t\cos{t} -t^2 + t +1 \]
i.e.
\[ e^t \left[(B-3A)\cos{t} + (-A-3B)\sin{t}\right]
-2Ct^2 -2(C+D)t + (2C -D -2E) = e^t\cos{t} -t^2 + t +1 
\]
i.e.
\begin{eqnarray*}
B-3A & = & 1 \\
-A-3B& = & 0 \\
-2C  & = & -1\\
-2(C+D) & = & 1 \\
2C -D -2E & = & 1
\end{eqnarray*}
and solving:
\[ A = -3/10\ ,\ B = 1/10\ ,\ C = 1/2\ ,\ D = -1\ ,\ E = 1/2 \]
and
\[ y_p(t) = 
-\frac{3}{10}e^t\cos{t} + \sfrac{1}{10}e^t\sin{t} +\sfrac{1}{2}t^2 -t 
+\sfrac{1}{2} \ . \]
\newpage
\section{Problem 4.8.36}
Determine the form of a particular solution to the DE
\[ y''+5y' + 6y = \sin{x} - \cos{2x} \ . \]
{\bf Solution:} \\
The characteristic equation is $r^2 +5r +6 = 0$ with roots $r =-2, -3$. Since
$i$ and $2i$ are not roots, try
\[ y_p(t) = A \cos{x}+B\sin{x}+C\cos{2x}+D\sin{2x} \ . \]
\section{Problem 4.8.45}
Find a particular solution to the higher order DE
\[ y''' - y'' + y = sin x \ . \]
{\bf Solution:} \\
The characteristic equation is $r^3 -r^2 +1 = 0$.
Since $i$ is not a root, the solution has the form
\[ y_p(t) = A \cos{x}+B\sin{x} \ . \]
\section{Problem 4.8.48}
Find a particular solution to the higher order DE
\[ y^{iv} - y'' - 8y = sin x \ . \]
{\bf Solution:} \\
The characteristic equation is $r^4 -r^2 -8 = 0$.
Since $i$ is not a root, the solution has the form
\[ y_p(t) = A \cos{x}+B\sin{x} \ . \]
\end{document}
