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{\bf Question}
\begin{description}
\item[(i)]
Find the general solution of the equation

$$\ds\frac{d^2y}{dx^2}+2\ds\frac{dy}{dx}-3y=0.$$

\item[(ii)]
Use the result of part (i) to find the solution of

$$\ds\frac{d^2y}{dx^2}+2\ds\frac{Dy}{dx}-3y=10 \sin x,$$

where $y=0$ and $\ds\frac{dy}{dx}=-5$ when $x=0$.
\end{description}

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$\ds\frac{d^2y}{dx^2}+2\ds\frac{dy}{dx}-3y=0$

Trial solution $y=Ae^{mx}$

$\Rightarrow$ Auxiliary equation of CF

$$m^2+2m-3=0$$

$\Rightarrow$ \begin{eqnarray*} m & = & \ds\frac{-2 \pm \sqrt{4-(4
\cdot 1 \cdot 3}}{2}\\ & = & \ds\frac{-2\pm 4}{2}\\ & = & 1\
\rm{or}\ -3 \end{eqnarray*}

Thus we have a solution

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

\newpage
\item[(ii)]
PI has form

$$y=A\sin x+B \cos x$$

Substitute into equation

\begin{eqnarray*} y' & = & A\cos x-B\sin x\\ y'' & = & -A\sin
x-B\cos x \end{eqnarray*}

$\Rightarrow -A\sin x-B\cos x+2A\cos x-2B \sin x-3A\sin x-3B\cos x
= 10\sin x$

$\Rightarrow \left. \begin{array} {rcl} -A-2B-3A & = & 10\\
-B+2A-3B & = & 0 \end{array} \right\} \left\{ \begin{array} {rcl}
-4A-2B & = & 10\\ 2A-4B & = & 0 \end{array} \right.$

$\Rightarrow A=2B$

\begin{eqnarray*} -4(2B)-2B & = & 10\\ -8B-2B & = & 10\\ B & = &
-1 \Rightarrow A=-2 \end{eqnarray*}

Therefore PI is

$$\un{y=-2\sin x-\cos x}$$

Whole solution is

\begin{eqnarray*} y & = & CF + PI\\ y & = & Ae^x+Be^{-3x}-2\sin
x-\cos x\\ y' & = & Ae^x-3Be^{-3x}-2\cos x+\sin x \end{eqnarray*}

Use $y(0)=0,\ y'(0)=-5$

$\left.\begin{array}{rcl} 0 & = & A+B-2\times0-1\\ -5 & = &
A-3B-2\times 1+0 \end{array} \right\}\left\{\begin{array}{rcl} 1 &
= & A+B\\ -3 & = & A-3B \end{array} \right.$

$\Rightarrow 4 = 4B \Rightarrow B=1$ Therefore $A=0$

Therefore  $$\un{y=e^{-3x}-2\sin x-\cos x}$$
\end{description}

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