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\begin{document}

\begin{center}
{\Large Math 2233}

The Laplace Transform Method for Nonhomogeneous ODEs: Example
\end{center}

\begin{eqnarray*}
y^{\prime \prime }+8y^{\prime }+15y &=&120e^{t} \\
y\left( 0\right)  &=&-7 \\
y^{\prime }\left( 0\right)  &=&53
\end{eqnarray*}

Taking the Laplace transform of the ODE yields%
\begin{equation*}
\left( s^{2}\mathcal{L}\left[ y\right] -sy\left( 0\right) -y^{\prime }\left(
0\right) \right) +8\left( \mathcal{L}\left[ y\right] -y\left( 0\right)
\right) +15\mathcal{L}\left[ y\right] =120\mathcal{L}\left[ e^{t}\right] 
\end{equation*}%
or%
\begin{equation*}
\left( s^{2}+8s+15\right) \mathcal{L}\left[ y\right] +7s-53+56=\frac{120}{s-1%
}
\end{equation*}%
or%
\begin{equation*}
\left( s+3\right) \left( s+5\right) \mathcal{L}\left[ y\right] =\frac{120}{%
s-1}-7s-3
\end{equation*}%
or%
\begin{eqnarray*}
\mathcal{L}\left[ y\right]  &=&\frac{120}{\left( s-1\right) \left(
s+3\right) \left( s+5\right) }-\frac{7s+3}{\left( s+3\right) \left(
s+5\right) } \\
&=&\frac{-7s^{2}+4s+123}{\left( s+5\right) \left( s-1\right) \left(
s+3\right) }
\end{eqnarray*}
: Now we have $\mathcal{L}\left[ y\right] $ expressed as a ratio of two
polynomials. \ Next, we set up a Partial Fractions Expansion:%
\begin{equation*}
\frac{-7s^{2}+4s+123}{\left( s+5\right) \left( s-1\right) \left( s+3\right) }%
=\frac{A}{s+5}+\frac{B}{s-1}+\frac{C}{s+3}
\end{equation*}%
We now need to complete the Partial Fractions Expansion to determine the
constants $A,B,C$. \ \ After multiplying both sides by $\left( s+5\right)
\left( s-1\right) \left( s+3\right) $, we get%
\begin{equation*}
-7s^{2}+4s+123=A\left( s-1\right) \left( s+3\right) +B\left( s+5\right)
\left( s+3\right) +C\left( s+5\right) \left( s-1\right) 
\end{equation*}%
This last equation has to be true for all $s$. \ Choosing some special,
convenient, values for $s$, shows%
\begin{eqnarray*}
s &=&-5\quad \Rightarrow \quad -7\left( 25\right) +4\left( -5\right)
+123=A\left( -6\right) \left( -2\right) +B\left( 0\right) +C\left( 0\right) 
\\
\quad  &\Rightarrow &\quad -72=12A\quad \Rightarrow \quad A=6
\end{eqnarray*}%
\begin{eqnarray*}
s &=&1\quad \Rightarrow \quad -7+4+123=A\left( 0\right) +B\left( 6\right)
\left( 4\right) +C\left( 0\right)  \\
\quad  &\Rightarrow &\quad 120=24B\quad \Rightarrow \quad B=5
\end{eqnarray*}%
\begin{eqnarray*}
s &=&-3\quad \Rightarrow \quad -7\left( 9\right) +4\left( -3\right)
+123=A\left( 0\right) +B\left( 0\right) +C\left( 2\right) \left( -4\right) 
\\
\quad  &\Rightarrow &\quad 48=-8C\quad \Rightarrow \quad C=-6
\end{eqnarray*}
Applying these values for $A,B,C$, 
\begin{eqnarray*}
\mathcal{L}\left[ y\right]  &=&\frac{-7s^{2}+4s+123}{\left( s+5\right)
\left( s-1\right) \left( s+3\right) }=\frac{A}{s+5}+\frac{B}{s-1}+\frac{C}{%
s+3} \\
&=&6\frac{1}{s+5}+5\frac{1}{s-1}-6\frac{1}{s+3} \\
&=&6\mathcal{L}\left[ e^{-5t}\right] +5\mathcal{L}\left[ e^{t}\right] -6%
\mathcal{L}\left[ e^{-3t}\right]  \\
&=&\mathcal{L}\left[ 6e^{-5t}+5e^{t}-6e^{-3t}\right] 
\end{eqnarray*}%
and so%
\begin{equation*}
y\left( t\right) =6e^{-5t}+5e^{t}-6e^{-3t}
\end{equation*}

\end{document}
