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\begin{document}

{\bf Question}

Solve the following initial value problems:

\begin{enumerate}

\item $9y''-12y'+4y=0 \quad y(0)=2 \quad y'(0)=-1$

\item $y''+4y'+3y=0 \quad y(0)=2 \quad y'(0)=-1$

\item $y''+4y'+5y=0 \quad y(0)=1 \quad y'(0)=0\qquad(*)$

\item $y''+4y'+4y=0 \quad y(-1)=2 \quad y'(-1)=1\qquad(*)$

\end{enumerate}



\vspace{0.25in}

{\bf Answer}

\begin{enumerate}
\item
We guess that the solution has the form $y=e^{mx}$ so that the
auxiliary equation is: $9m^2-12m+4 =0$\\ which has roots $\ds
m=\frac{12\pm\sqrt{144-144}}{18}= \frac{2}{3},\frac{2}{3}$\\ As
the root is repeated the general solution is: \\$\ds y= Ae^{2
x/3}+ Bxe^{2 x/3} = (A+Bx)e^{2x/3}$\\ Let $y(0)=2$ so that $A=2$\\
Let $y'(0)=-1$ so that using $y'= 2A/3e^{2 x/3}+ 2Bx/3e^{2 x/3}+
Be^{2 x/3}$\\ we require $2A/3+B=-1$\\ Hence $A=2$, $B=-7/3$ and
the solution is:\\ $$ (2-7x/3)e^{2x/3} $$

\item
Auxiliary equation is: $m^2+4m+3 =0$\\ which has roots $\ds
m=\frac{-4\pm\sqrt{16-12}}{2}=-1,-3$\\ The general solution is:
$\ds y= Ae^{- x}+ Bxe^{-3 x}$\\ Let $y(0)=2$ so that $A+B=2$\\ Let
$y'(0)=-1$ so that $-A-3B=-1$\\ Hence $A=5/2$, $B=-1/2$ and the
solution is:\\ $$ y=\frac{5}{2}e^{-x}-\frac{1}{2}e^{-3x} $$

\item
Auxiliary equation is: $m^2+4m+5 =0$\\ which has roots $\ds
m=\frac{-4\pm\sqrt{16-20}}{2}=-2+i,-2-i$\\ The general solution
is: $\ds y= e^{-2x}(A\cos x + B\sin x)$\\ Let $y(0)=1$ so that
$A=1$\\ Let $y'(0)=0$ so that $-2A+B=0$\\ Hence $A=1$, $B=2$ and
the solution is:\\ $$ y= e^{-2x}(\cos x + 2\sin x) $$

\item
Auxiliary equation is: $m^2+4m+4 =0$\\ which has roots $\ds
m=\frac{-4\pm\sqrt{16-16}}{2}=-2,-2$\\ As the roots are repeated
the general solution is:\\ $\ds y= e^{-2x}(A+ Bx)$\\ Let $y(-1)=2$
so that $e^{2}(A-B)=2$\\ Let $y'(-1)=1$ so that
$-2e^{2}(A-B)+e^{2}B=1$\\ Hence $A=7e^{-2}$, $B=5e^{-2}$ and the
solution is:\\ $$ y= e^{-2x}(7e^{-2}+5xe^{-2})=(5x+7)e^{(-2(x+1))}
$$

\end{enumerate}


\end{document}
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