\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\newcommand{\lin}{\int_0^\infty}
\parindent=0pt
\begin{document}


{\bf Question}

Find the Laplace transform of each of the following functions. In
each case specify the values of $s$ for which the transform
exists.
\begin{description}
\item[(a)] $6\sin2t-5\cos2t$
\item[(b)] $(\sin t-\cos t)^2$
\item[(c)] $(t^2+1)^2$
\item[(d)] $t^3e^{-2t}$
\item[(e)] $e^{-4t}\cosh2t$
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds \mathcal{L} \{ 6 \sin 2t-5 \cos 2t \} = \int_0^\infty e^{-st}
\{ 6\sin 2t-5\cos 2t \} \, dt$. Now
\begin{eqnarray*}
I  = \lin \sin 2 t dt & = & \left[ -{1 \over s} e^{-st} \sin 2t
\right]_0^\infty + \lin {2 \over s} \cos 2 t \, dt \\ &
 = & 0 + \left[ -{2 \over {s^2}} e^{-st} \cos 2 t \right]_0^\infty + \lin {-4
\over {s^2}} \sin 2t \, dt \\ & = & {2 \over {s^2}} - {4 \over
{s^2}}I \end{eqnarray*} Hence $\ds {{s^2+4} \over {s^2}}I={2 \over
{s^2}}$, which gives $\ds I={2 \over {s^2+4}}$.

Similarly $\ds\lin \cos 2t dt= {s \over {s^2+4}}$, and hence $\ds
\mathcal{L}\{6\sin 2t-5\cos 2t\}={{12-5s} \over {s^2+4}}$ for
$s>0$.

\item[(b)]
\begin{eqnarray*} \mathcal{L} \{ (\sin t - \cos t)^2 \} & = & \mathcal{L}
\{\sin^2 t-2 \sin t \cos t+\cos^2 t \} \\ & = & \mathcal{L} \{1 -
\sin 2 t \} \\ & = & {1 \over s} - {2 \over {s^2+4}} \quad s>0
\end{eqnarray*}

\item[(c)]
$\ds \mathcal{L}\{(t^2+1)^2\}=\mathcal{L}\{t^4+2t^2+1\}$. Now $\ds
\mathcal{L}\{t^n\}={{n!} \over {s^{n+1}}}$. So that

$\ds \mathcal{L}\{(t^2+1)^2\}={24 \over {s^5}}+{4 \over {s^3}}+{1
\over s} ={{24+4s^2+s^4} \over {s^5}}$, for $s>0$.

\item[(d)]
$\ds \mathcal{L}\{t^3e^{-2t}\}=F(s+2)$ where $\ds
F(s)=\mathcal{L}\{t^3\}={6 \over {s^4}}$.

Hence $\ds \mathcal{L}\{t^3e^{-2t}\}={6 \over {(s+2)^4}}$, for
$s>-4$.

\item[(e)]
$\ds \mathcal{L}\{e^{-4t}\cosh2t\}=F(s+4)$ where $\ds
F(s)=\mathcal{L}\{\cosh2t\}={s \over {s^2-4}}$.

Hence $\ds \mathcal{L}\{e^{-4t}\cosh2t\}={{s+4} \over
{s^2+8s+12}}$, for $s>-4$.
\end{description}



\end{document}