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QUESTION

\begin{description}

\item[(i)]
Using partial fractions and the table of inverse Laplace
transforms show that

$${\cal{L}}^{-1}\left\{\frac{1}{(s+2)(s+1)^2}\right\}=e^{-2t}-e^{-t}+te^{-t}.$$

\item[(ii)]
Use Laplace transforms and part (i) to find the solution of

$$\frac{d^2x}{dt^2}+3\frac{dx}{dt}+2x=2e^{-t}$$

which satisfies the conditions $x=0$ and $\ds\frac{dx}{dt}=0$ when
$t=0$.

What is the behaviour of the solution $x$ and $\ds\frac{dx}{dt}$
as $t\to\infty$?

\end{description}



\bigskip

ANSWER

\begin{description}

\item[(i)]
\begin{eqnarray*}
&&\frac{1}{(s+2)(s+1)^2}
=\frac{A}{s+2}+\frac{B}{s+1}+\frac{C}{(s+1)^2}\\
&&=\frac{A(s+1)^2+B(s+1)(s+2)+C(s+2)}{(s+2)(s+1)^2}
\end{eqnarray*}
 therefore
$A(s+1)^2+B(s+1)(s+2)+C(s+2)=1$

\begin{tabular}{lll}
$s=-1$&$0+0+C(1)=1,$&$C=1$\\
$s=-2$&$A(-1)^2+0+0=1$&$A=1$\\coefficient of
$s^2$&$A+B=0,$&$B=-A=-1$
\end{tabular}

Therefore

$\ds{\cal{L}}^{-1}\left\{\frac{1}{(s+2)(s+1)^2}\right\}\\
={\cal{L}}^{-1}\left\{\frac{1}{s+2}-\frac{1}{s+1}+\frac{1}{(s+1)^2}\right\}=e^{-2t}-e^{-t}+te^{-t}$

\item[(ii)]
$\ds\frac{d^2x}{dt^2}+3\frac{dx}{dt}+2x=2e^{-t},\hspace{1cm}x(0)=\frac{dx}{dt}(0)=0$

Taking the Laplace transform

\begin{eqnarray*}
\left(s^2X-sx(0)-\frac{dx}{dt}(0)\right)+3(sX-x(0))+2X&=&\frac{2}{s+1}\\
s^2X+3sX+2X&=&\frac{2}{s+1}\\ (s^2+3s+2)X&=&\frac{2}{s+1}\\
(s+2)(s+1)X&=&\frac{2}{s+1}\\ X&=&\frac{2}{(s+2)(s+1)^2}
\end{eqnarray*}

Hence $\ds
x(t)={\cal{L}}^{-1}\left\{\frac{2}{(s+2)(s+1)^2}\right\}=2(e^{-2t}-e^{-t}+te^{-t})$\\
i.e. $\ds
x(t)=2\left(\frac{1}{e^{2t}}-\frac{1}{e^t}+\frac{t}{e^t}\right)\to0$
as $t\to\infty$.

\begin{eqnarray*}
\frac{dx}{dt}&=&2\{-2e^{-2t}+e^{-t}+e^{-t}-te^{-t}\}\\
&=&2\{-2e^{-2t}+2e^{-t}-te^{-t}\}\\
&=&2\left(-\frac{2}{e^{2t}}+\frac{2}{e^t}-\frac{t}{e^t}\right)\\
&\to&0 {\rm\ as\ } t\to\infty
\end{eqnarray*}



\end{description}




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