\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: Laplace}

Write down the Laplace transforms of the following functions
$$\mathrm{(i)}\hspace{0.2in} \sin3t+t^2\hspace{0.5in}
\mathrm{(ii)}\hspace{0.2in}\left(1+\mathrm{e}^{-t}\right)^2\hspace{0.5in}
\mathrm{(iii)}\hspace{0.2in} t\mathrm{cosh}t.$$

Find the inverse transform of
$$\frac{6p^2\mathrm{e}^{-p}}{(p^2-1)(p^2+2)}$$ \vspace{0.5in}

{\bf Solution}

$\displaystyle \mathrm{(i)}\ \
L(\sin3t+t^2)=\frac{3}{p^2+9}+\frac{2}{p^3}.$

$\displaystyle \mathrm{(ii)\ \
}L\left((1+\mathrm{e}^{-t})^2\right)=L(1+2\mathrm{e}^{-t}+\mathrm{e}^{-2t})
=\frac{1}{p}+\frac{2}{p+1}+\frac{1}{p+2}.$

$\displaystyle \mathrm{(iii)}\ \ L(t\cosh t)=L\left(\frac{t\,
\mathrm{e}^t+t\, \mathrm{e}^{-t}}{2
}\right)=\frac{1}{2(p-1)^2}+\frac{1}{p+2}.$ \vspace{0.3in}

$$\frac{6p^2\mathrm{e}^{-p}}{(p^2-1)(p^2+2)}
=\mathrm{e}^{-p}\left(\frac{2}{p^2-1}+\frac{4}{p^2+2}\right) =
\left(\frac{1}{p-1}+\frac{1}{p+1}+\frac{4}{p^2+2}\right)\mathrm{e}^{-p}$$

Inverting using the second shift theorem gives

$$L^{-1}\left(\frac{6p^2\mathrm{e}^{-p}}{(p^2-1)(p^2+2)}\right)
=\left(\mathrm{e}^{(t-1)}-\mathrm{e}^{-(t-1)}+
\frac{4}{\sqrt2}\sin\left(\sqrt2(t-1)\right)\right)H(t-1).$$

\end{document}