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{\bf Exam Question

Topic: Laplace}

In this question $L(y(x))$ denotes the Laplace transform of
$y(x).$

(i) Show that $\displaystyle L\left(\mathrm{e}^{-ax}\cos
bx\right)=\frac{p+a}{(p+a)^2+b^2}$ for $p&gt;-a.$

(ii) Show that $\displaystyle L\left(\mathrm{e}^{-ax}\sin
bx\right)=\frac{b}{(p+a)^2+b^2}$ for $p&gt;-a.$

(iii) Find $ y(x)$ if $\displaystyle L(y(x))=\frac{p}{p^2-4p+13}.$
 \vspace{0.5in}

{\bf Solution} $$\mbox{Let}\ \ I_1=L\left(\mathrm{e}^{-ax}\cos
bx\right) =\int_0^{\infty}\mathrm{e}^{-(p+a)x}\cos bx\, dx$$

$$\mbox{Let}\ \ I_2=L\left(\mathrm{e}^{-ax}\sin bx\right)
=\int_0^{\infty}\mathrm{e}^{-(p+a)x}\sin bx\, dx$$

\begin{eqnarray}
bI_1&amp;=&amp;\left[\mathrm{e}^{-ax}\sin
bx\right]_0^{\infty}+(p+a)I_2=(p+a)I_2 \nonumber \\&amp;\Rightarrow&amp;
bI_1-(p+a)I_2=0
\\ -bI_2&amp;=&amp;\left[\mathrm{e}^{-ax}\cos
bx\right]_0^{\infty}+(p+a)I_1=-1+(p+a)I_1 \nonumber
\\&amp;\Rightarrow&amp; (p+a)I_1+bI_2=1
\end{eqnarray}
\begin{eqnarray*}
&amp;&amp;b\times (1)+(p+a)\times (2)\ \ \mbox{gives}\ \
\left[(p+a)^2+b^2\right]I_1=(p+a)\\&amp;&amp;\Rightarrow
I_1=\frac{p+a}{(p+a)^2+b^2}\ \ \mbox{and so}\ \
I_2=\frac{b}{(p+a)^2+b^2}
\end{eqnarray*}
\begin{eqnarray*}
L(y(x))&amp;=&amp;\frac{p}{p^2-4p+13}=\frac{p-2}{(p-2)^2+3^2}+\frac{2}{(p-2)^2+3^2},\\
\mbox{so}\ \ y(x)&amp;=&amp;\mathrm{e}^{2x}\left(\cos
3x+\textstyle{\frac{2}{3}}\sin 3x\right).
\end{eqnarray*}


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