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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Solve the equations

\begin{description}
\item[(i)]
$\ds\frac{d^2y}{dx^2}-5\ds\frac{dy}{dx}+6y=0$

\item[(ii)]
$4\ds\frac{d^2y}{dx^2}-12\ds\frac{dy}{dx}+9y=0$

\item[(iii)]
$\ds\frac{d^2y}{dx^2}+2\ds\frac{dy}{dx}+2y=0$
\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)] %note ref to class notes%
$\ds\frac{d^2y}{dx^2}-5\ds\frac{dy}{dx}+6y=0$

Try solution

$\left\{\begin{array}{ccl} y & = & Ae^{kx}\\ \ds\frac{dy}{dx} & =
& Ake^{kx}\\ \ds\frac{d^2y}{dx^2} & = & Ak^2e^{kx}\end{array}
\right.$

$\Rightarrow$ auxiliary equation

$\ k^2-5k+6=0$

$\Rightarrow (k-2)(k-3)=0$

$\Rightarrow k=2$ or $k=3$

Unequal real roots.  Thus the \un{general} solution is (see notes)

$$\un{y=Ae^{2x}+Be^{3x}}$$

$A,\ B$ arbitrary constants

\newpage
\item[(ii)]
$4\ds\frac{d^2y}{dx^2}-12\ds\frac{dy}{dx}+9y=0$

Try solution

$\left\{\begin{array}{ccl} y & = & Ae^{kx}\\ \ds\frac{dy}{dx} & =
& Ake^{kx}\\ \ds\frac{d^2y}{dx^2} & = & Ak^2e^{kx}\end{array}
\right.$

$\Rightarrow$ auxiliary equation

$\ 4k^2-12k+9=0$

$\Rightarrow (2k-3)^2=0$

$\Rightarrow k=\ds\frac{3}{2},\ \ds\frac{3}{2}$

Two equal real roots.  Thus the \un{general} solution is (see
notes)

$$\un{y=(A+Bx)e^{\frac{3x}{2}}}$$

\item[(iii)]
$\ds\frac{d^2y}{dx^2}+2\ds\frac{dy}{dx}+2y=0$

Try solution

$\left\{\begin{array}{ccl} y & = & Ae^{kx}\\ \ds\frac{dy}{dx} & =
& Ake^{kx}\\ \ds\frac{d^2y}{dx^2} & = & Ak^2e^{kx}\end{array}
\right.$

$\Rightarrow$ auxiliary equation is

$\ k^2+2k+2=0$

$\Rightarrow k=\ds\frac{-2 \pm\sqrt{4-4\cdot
2}}{2}=\ds\frac{-2\pm\sqrt{-4}}{2}=-1\pm i$

This is an imaginary roots case.

Thus $y=Ae^{(-1+i)}x+Be^{(-1-i)}x=\un{e^{-x}(Ae^{ix}+Be^{-ix})}$

Alternatively $\un{y=e^{-x}(C \cos x+D \sin x)}$

Probably better since this is obviously real, like the equation.
\end{description}

\end{document}