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\begin{document}


{\bf Question}

Use Laplace transforms to solve the following systems of
equations:
\begin{description}
\item[(a)] \begin{eqnarray*} x' & = & y+e^t \\ y' & = & -2x+3y+2 \end{eqnarray*}
with initial conditions $x(0)=2$, $y(0)=2$.
\item[(b)] \begin{eqnarray*} x' & = & 2x+y+te^{2t} \\ y' & = & -4x+2y-e^{2t}\end{eqnarray*}
with initial conditions $x(0)=1$, $y(0)=1$.
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]

Taking the Laplace transform of the equations gives:
\begin{eqnarray} sX-2 & = & Y+{1 \over {s-1}} \\ sY-2 & = & -2X+3Y+{2 \over s}
 \end{eqnarray} From (1) we get $\ds X={Y \over s}+{1 \over
{s(s-1)}}+{2 \over s}$

Substituting in (2) gives $\ds sY-2=-{{2Y} \over s}-{2 \over
{s(s-1)}}-{4 \over s}+3Y+{2 \over s}$. Rearranging and multiplying by
$s^2$ gives

$\ds (s^2-3s+2)Y=2s-{2 \over {s-1}}-2$ which gives

$\ds Y={2 \over {s-2}}-{2 \over {(s-1)^2(s-2)}}$. On taking partial
fractions this simplifies to

$\ds Y={2 \over {s-1}}+{2 \over {(s-1)^2}}$. Hence $y(t)=2e^t+2te^t$.

We now rewrite the second differential equation as
$\ds x={1 \over 2}(3y+2-y')$ and substitute for $y(t)$ to obtain

$x(t)=e^t+2te^t+1$.

\item[(b)] Taking the Laplace transform of the equations gives:
\begin{eqnarray*} sX-1 &  = & 2X+Y+{1 \over {(s-2)^2}}
\hspace{1.25in}(1)
\\ sY-1 & = & -4X+2Y-{1 \over {(s-2)}} \hspace{1.1in}(2)
\end{eqnarray*} Rearranging gives \begin{eqnarray*} (s-2)X-Y & = &
1+{1 \over {(s-2)^2}} \hspace{1.2in}(3) \\ 4X+(s-2)Y & = & 1-{1
\over {(s-2)}} \hspace{1.25in}(4)
\end{eqnarray*}

$(s-2)(3)+(4) \quad \Rightarrow \quad [4+(s-2)^2]X=1+s-2$

$\ds \Rightarrow\quad X={1 \over 2}{2 \over {(s-2)^2+4}}+{{s-2}
\over {(s-2)^2+4}}$

$\ds \Rightarrow\quad x(t)={1 \over 2}e^{2t}\sin 2t
+e^{2t}\cos2t$.

But from the first differential equation $y=x'-2x-te^{2t}$ and
substituting for $x(t)$ gives

$y(t)=-2e^{2t}\sin2t+e^{2t}\cos2t-te^{2t}$.
\end{description}


\end{document}
