<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\newcommand{\lin}{\int_0^\infty}
\parindent=0pt
\begin{document}


{\bf Question}

Solve each of the following second order equations by using
Laplace transforms and check your answers.
\begin{description}
\item[(a)] $y''+4y=9t \qquad y(0)=0, \qquad y'(0)=7$
\item[(b)] $y'''+y=e^t \qquad y(0)=0, \qquad y'(0)=0, \qquad y''(0)=0$
\item[(c)] $y''+4y=1-H(t-1) \qquad y(0)=0, \qquad y'(0)=1$
\item[(d)] $y''+4y=\delta(t-2) \qquad y(0)=0, \qquad y'(0)=1$
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds y''+4y=9t$, $y(0)=0$, $y'(0)=7$. Taking the Laplace transform
gives

$\ds s^2Y-sy(0)-y'(0)+4Y={9 \over {s^2}} \quad \Rightarrow \quad
(s^2+4)Y={9 \over {s^2}}+7$ \begin{eqnarray*} \Rightarrow \quad Y
&amp;= &amp; {9 \over {s^2(s^2+4)}}+{7 \over {(s^2+4)}} \\ &amp;= &amp; {9 \over
4}{1 \over {s^2}}-{9 \over 4}{1 \over {(s^2+4)}}+{7 \over
{(s^2+4)}} \\ &amp;= &amp; {9 \over 4}{1 \over {s^2}}+{19 \over 8}{2 \over
{s^2+4}} \end{eqnarray*} $\ds \Rightarrow \quad y(t)={{9t} \over
4}+{19 \over 8}\sin 2t$.

\item[(b)]
$\ds y'''+y=e^t$, $y(0)=0$, $y'(0)=0$, $y''(0)=0$. Taking the
Laplace transform gives

$\ds s^3Y-s^2y(0)-sy'(0)-y''(0)+Y={1 \over {s-1}}, \quad
\Rightarrow \quad (s^3+1)Y={1 \over {s-1}}$. Hence

$\ds Y={1 \over {(s^3+1)(s-1)}}={1 \over {(s+1)(s^2-s+1)(s-1)}}
={A \over {(s+1)}}+{B \over {(s-1)}}+{{Cs+D} \over {s^2-s+1}}$.

Hence $A(s-1)(s^2-s+1)+B(s+1)(s^2-s+1)+(Cs+D)(s+1)(s-1)=1$.

$\Rightarrow \quad (A+B+C)s^3+(-2A+D)s^2+(2A-C)s+(-A+B-D)=1$.

Thus $A+B+C=0$, $-2A+D=0$, $2A-C=0$ and $-A+B-D=1$. Solving these
equations gives

$\ds A=-{1 \over 6}, \quad B={1 \over 2}, \quad C=-{1 \over 3},
\quad D=-{1 \over 3}$. So that \begin{eqnarray*} Y(s) &amp;= &amp;  -{1
\over 6}{1 \over {(s+1)}}+{1 \over 2}{1 \over {(s-1)}} -{1 \over
3}{{s+1} \over {s^2-s+1}} \\ &amp; = &amp; -{1 \over 6}{1 \over
{(s+1)}}+{1 \over 2}{1 \over {(s-1)}} -{1 \over 3}{{(s-1/2)+3/2}
\over {(s-1/2)^2+(\sqrt3/2)^2}} \\ &amp; = &amp; -{1 \over 6}{1 \over
{(s+1)}}+{1 \over 2}{1 \over {(s-1)}} \\ &amp; &amp; -{1 \over
3}{{(s-1/2)} \over {(s-1/2)^2+(\sqrt3/2)^2}} -{1 \over
\sqrt3}{{\sqrt3/2} \over {(s-1/2)^2+(\sqrt3/2)^2}} \end{eqnarray*}

So that $\ds y(t)= -{1 \over 6}e^{-t}+{1 \over 2}e^t
-{1 \over 3}e^{1/2t}\cos{{\sqrt 3} \over 2}t
-{1 \over {\sqrt3}}e^{1/2t}\sin{{\sqrt 3} \over 2}t$.

\item[(c)]
$\ds y''+4y=1-H(t-1)$, $y(0)=0$, $y'(0)=1$. Taking the Laplace
transform gives

$\ds s^2Y-sy(0)-y'(0)+4Y={1 \over s}-{{e^{-s}} \over s}$ Hence

$\ds (s^2+4)Y={1 \over s}-{{e^{-s}} \over s}+1$. Thus
\begin{eqnarray*} Y &amp; = &amp; {1 \over {s(s^2+4)}}-{{e^{-s}} \over
{s(s^2+4)}}+{1 \over {s(s^2+4)}} \\ &amp; = &amp; {1 \over 4}{1 \over
s}-{1 \over 4}{s \over {s^2+4}} +{1 \over 2}{2 \over {s^2+4}} -{1
\over 4}{{e^{-s}} \over s}+{1 \over 4}{{se^{-s}} \over {s^2+4}}
\end{eqnarray*} $\ds \Rightarrow \quad y(t)={1 \over 4}-{1 \over 4}\cos2t+{1
\over 2}\sin2t -{1 \over 4}H(t-1)+{1 \over 4}H(t-1)\cos2(t-1)$.

\item[(d)]
$\ds y''+4y=\delta(t-2)$, $y(0)=0$, $y'(0)=1$. Taking the Laplace
transform gives

$\ds s^2Y-sy(0)-y'(0)+4Y=e^{-2s}$, Hence $(s^2+4)Y=1+e^{-2s}$.

$\ds \Rightarrow \quad Y={1 \over 2}{2 \over {s^2+4}}+{{e^{-2s}}
\over 2}{2 \over {s^2+4}}$.

$\ds \Rightarrow \quad y(t)={1 \over 2}\sin2t+{1 \over
2}H(t-2)\sin2(t-2)$.
\end{description}



\end{document}
</pre></body></html>