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\begin{document}

{\bf Question}

Find the general solution of the following differential equations:

\begin{enumerate}

\item $y''+5y'=0$

\item $y''-6y'+9y=0 \qquad(*)$

\item $y''+2y'-8y=0 \qquad(*)$

\item $y''+6y'+13y=0\qquad(*)$

\end{enumerate}


\vspace{0.25in}

{\bf Answer}

Auxiliary equation is: $m^2+5m =0$\\ which has roots $m=-5,0$\\
Hence general solution is: $\ds y= Ae^{-5 x}+ Be^{0 x} = Ae^{-5
x}+ B $
\begin{enumerate}
\item
Auxiliary equation is: $m^2-6m+9 =0$\\ which has roots $m=3,3$.\\
Since the roots are repeated the general solution is:\\ $\ds y=
Ae^{3 x}+ Bxe^{3 x} = (A+Bx)e^{3 x} $
\item
Auxiliary equation is: $m^2+2m-8 =0$\\ which has roots $\ds
m=\frac{-2\pm\sqrt{4+32}}{2}=-4,2$\\ Hence general solution is:
$\ds y= Ae^{-4 x}+ Be^{2 x}$
\item
Auxiliary equation is: $m^2+6m+13 =0$\\ which has roots $\ds
m=\frac{-6\pm\sqrt{36-52}}{2}=-3+2i,-3-2i$\\ Hence the general
solution is:\\ $\ds y= Ae^{(-3+2i) x}+ Be^{(-3-2i) x} =
e^{-3x}\Big(C\cos(2x)+D\sin(2x)\Big)$

\end{enumerate}




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