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{\bf Exam Question

Topic: LaplaceODE}

Suppose that
\begin{eqnarray*}
5y_1''-10y_1'-13y_2'+11y_1+13y_2&=&0\\
y_2''+13y_1'-2y_2'-13y_1-29y_2&=&0,
\end{eqnarray*}
and that\ \  $y_1(0)=1,\ \ y_1'(0)=4,\ \ y_2(0)=1,\ \ y_2'(0)=3.$

Show that \ $\displaystyle
\bar{y}_1=\frac{p^3+6p-19}{(p^2-2p-3)(p^2-2p+10)}$ \ and hence
find $y_1(x).$

\vspace{0.5in}

{\bf Solution}

Transforming the differential equations gives
\begin{eqnarray}
&&5(p^2\bar{y}_1-p-4)-10(p\bar{y}_1-1)-13(\bar{y}_2-1)+11\bar{y}_1+13\bar{y}_2=0
\nonumber\\ &&\Rightarrow
(5p^2-10p+11)\bar{y}_1+(-13p+13)\bar{y}_2=5p-3\\
&&(p^2\bar{y}_2-p-3)+13(p\bar{y}_1-1)-2(p\bar{y}_2-1)-13\bar{y}_1-29\bar{y}_2=0
\nonumber\\
&&\Rightarrow(13p-13)\bar{y}_1+(p^2-2p-29)\bar{y}_2=p+14
\end{eqnarray}
Now\ \ $(p^2-2p-29)\times (1)-(-13p+13)\times (2)$\ \ gives

\begin{eqnarray*}
&&5(p^4-4p^3+11p^2-14p-30)\bar{y}_1=5(p^3+6p-19)\\ &&\Rightarrow
\bar{y}_1=\frac{p^3+6p-19}{(p^2-2p-3)(p^2-2p+10)}\\ &&\Rightarrow
\bar{y}_1=\frac{p-1}{(p-1)^2-2^2}+\frac{3}{(p-1)^2+3^2}\\
&&\Rightarrow y_1(x)=\mathrm{e}^{-x}(\cosh 2x+\sin 3x).
\end{eqnarray*}


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