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QUESTION

Use Laplace transforms to solve the simultaneous equations

\begin{eqnarray*}
y'+y+z'&amp;=&amp;0\\ y'-y+2z'&amp;=&amp;e^{-x}
\end{eqnarray*}

given that $y=\frac{1}{2}$ and $z=0$ at $x=0$.

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ANSWER

 $\left.\begin{array}{r}y'+y+z'=0\\y'-y+2z'=e^{-x}\end{array}\right\}\
 \ y(0)=\frac{1}{2},\ \ z(0)=0$
 $$
 \begin{array}{l}sY(s)-y(0)+Y(s)+sZ(s)-z(0)=0\\sY(s)-y(0)-Y(s)+2(sZ(s)-z(0))=\frac{1}{s+1}
 \end{array}$$

 \begin{eqnarray}(s+1)Y+sZ=\frac{1}{2}\\(s-1)Y+2sZ=\frac{1}{2}+\frac{1}{s+1}
 \end{eqnarray}
 $2\times(1)-(2)\Rightarrow (s+3)Y=\frac{1}{2}-\frac{1}{s+1}$

 \begin{eqnarray*}
 Y&amp;=&amp;\frac{1}{2(s+3)}-\frac{1}{(s+1)(s+3)}\\
 &amp;=&amp;\frac{1}{s+3}-\frac{1}{2}\frac{1}{s+1}\\
 &amp;\Rightarrow&amp;
 y=e^{-3x}-\frac{1}{2}e^{-x}\\
&amp;\Rightarrow&amp;
 Z=\frac{1}{2s}-\frac{s+1}{s}Y\\
 &amp;=&amp;\frac{1}{2s}-\frac{s+1}{s(s+3)}+\frac{1}{2s}\\
 &amp;=&amp;\frac{1}{s}-\frac{1}{s+3}-\frac{1}{s(s+3)}\\
 &amp;=&amp;\frac{2}{3}\frac{1}{s+3}\\
 &amp;\Rightarrow&amp;
 z=\frac{2}{3}=\frac{2}{3}e^{-3x}
\end{eqnarray*}


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