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

\section{Problem 7.2.15}
Determine the transform
\[ {\cal L}\left\{ t^3 - t e^t + e^{4t} \cos{t} \right\} \]
{\bf Solution:}\\
\begin{eqnarray*}
 {\cal L}\left\{ t^3 - t e^t + e^{4t} \cos{t} \right\} & = & 
 {\cal L}\left\{ t^3\right\} - {\cal L}\left\{t e^t\right\} + {\cal L}\left\{e^{4t} \cos{t} \right\} \\
& = & \frac{3!}{s^{3+1}} - \frac{1}{(s-1)^{1+1}} + \frac{s-4}{(s-4)^2 + 1} \\
& = & \frac{6}{s^{4}} - \frac{1}{s^{2}-2s+1} + \frac{s-4}{s^2-8s + 17}
\end{eqnarray*}

\section{Problem 7.3.5}
Determine the transform
\[ {\cal L}\left\{ 2 t^2 e^{-t} - t  +  \cos{4t} \right\} \]
{\bf Solution:}\\
\begin{eqnarray*}
{\cal L}\left\{ 2 t^2 e^{-t} - t  +  \cos{4t} \right\} & = &
2{\cal L}\left\{t^2 e^{-t}\right\} -{\cal L}\left\{ t\right\}  +  {\cal L}\left\{\cos{4t} \right\} \\
& = & 
2 \frac{2!}{(s+1)^{2+1}} -\frac{1!}{s^{1+1}}  +  \frac{s}{s^2+4^2} \\
& = & 
 \frac{4}{(s+1)^{3}} -\frac{1}{s^{2}}  +  \frac{s}{s^2+16}
\end{eqnarray*}
\newpage

\section{Problem 7.3.19}
Determine the transform
\[ {\cal L}\left\{ \cos{nt}\sin{mt}\right\}\ \ ,\ \ m \ne n  \]
{\bf Solution:} \\
Using the familiar trigonometric identity:
\begin{eqnarray*}
\sin{\left(A\pm B\right)} & = & \sin{A}\cos{B}\pm\cos{A}\sin{B} \\
2\sin{A}\cos{B} & = & \sin{\left(A-B\right)} + \sin{\left(A+B\right)} \\
2\cos{A}\sin{B} & = & \sin{\left(A+B\right)} - \sin{\left(A-B\right)}
\end{eqnarray*}
so that, if we set 
\[ A = nt\ ,\  B = mt \]
i.e.
\[ A+B=(n+m)t\ ,\  A-B=(n-m)t \]
we have
\[
\cos{nt}\sin{mt} = \frac{1}{2}\sin{\left(n+m\right)t} 
- \frac{1}{2}\sin{\left(n-m\right)t}
\]
Applying these identities to our problem:
\begin{eqnarray*}
{\cal L}\left\{ \cos{nt}\sin{mt}\right\}
& = & \frac{1}{2}{\cal L}\left\{ \sin{\left(n+m\right)t}\right\}
- \frac{1}{2} {\cal L}\left\{ \sin{\left(n-m\right)t} \right\} \\
& = & \frac{1}{2} \frac{n+m}{s^2 + (n+m)^2}
- \frac{1}{2} \frac{n-m}{s^2 + (n-m)^2}
\end{eqnarray*}

\newpage

\section{Problem 7.3.25}
Use the formula
\[ {\cal L}\left\{ t^n f(t) \right\} = (-1)^n \frac{ d^n F(s)}{ds^n} (s) \ .\]
to help determine:
\begin{enumerate}
\item ${\cal L}\left\{t \cos{bt} \right\} $
\item ${\cal L}\left\{t^2 \cos{bt} \right\} $
\end{enumerate}
{\bf Solution:} \\
Since
\[ F(s) :=  {\cal L}\left\{ \cos{bt} \right\} = \frac{s}{s^2+b^2} \]
we have
\begin{eqnarray*}
{\cal L}\left\{t \cos{bt} \right\} & = & - \frac{ dF(s)}{ds} \\
& = &-\frac{ d}{ds} \frac{s}{s^2+b^2} \\
& = &-\frac{1}{s^2+b^2} + \frac{2s^2}{\left(s^2+b^2\right)^2} \\
& = &-\frac{\left(s^2+b^2\right) - 2s^2}{\left(s^2+b^2\right)^2} \\
& = & \frac{s^2 - b^2}{\left(s^2+b^2\right)^2}
\end{eqnarray*}
\begin{eqnarray*}
{\cal L}\left\{t \cos{bt} \right\} & = & \frac{ d^2F(s)}{ds^2} \\
& = &-\frac{ d}{ds} \frac{s^2 - b^2}{\left(s^2+b^2\right)^2}  \\
& = &-\frac{2s}{\left(s^2+b^2\right)^2}  
+\frac{4s(s^2 - b^2)}{\left(s^2+b^2\right)^3}  \\
& = &-\frac{2s\left(s^2+b^2\right)}{\left(s^2+b^2\right)^3}  
+\frac{4s(s^2 - b^2)}{\left(s^2+b^2\right)^3}  \\
& = & \frac{2s(s^2 - 3 b^2)}{\left(s^2+b^2\right)^3}
\end{eqnarray*}

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