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{\bf Question}

Solve the following differential equations, using the specified
boundary conditions.
\begin{description}
\item[(a)]
$\frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=0$, where $y = 2$ and
$\frac{dy}{dx} = -3$ when $x=0$
\item[(b)]
$\frac{d^2y}{dx^2}+6\frac{dy}{dx}+13y=0$, where $y = 1$ and
$\frac{dy}{dx} = 2$ when $x=0$
\item[(c)]
$\frac{d^2y}{dx^2}-6\frac{dy}{dx}+9y=0$, where $y = 2$ and
$\frac{dy}{dx} = 4$ when $x=0$
\end{description}


{\bf Answer}

\begin{description}
\item[(a)]

Auxiliary equation: $\lambda^2 + 3\lambda + 2  =  0$

Factorises to give: $(\lambda + 1)(\lambda +2)  =  0$

so two distinct real roots $\lambda_1 = -1$ and $\lambda_2 = -2$
\begin{eqnarray*}
{\rm General\  solution:\ } y & = & Ae_{-x} +B e^{-2x} \\
\frac{dy}{dx} & = & -Ae^{-x} -2e^{-2x} \\ {\rm Boundary\
conditions:\ } 2 & = & A + B \\ -3 & = & -A -2B \\ \Rightarrow B &
 = & 1 {\rm \ and\ } A=1 \\
{\rm Particular\  solution:\ } y & = & e^{-x} + Be^{-2x}
\end{eqnarray*}

\item[(b)]

Auxiliary equation: $\lambda^2 + 6\lambda + 13  =  0$

$\lambda_1,\lambda_2  =  \frac{-6 \pm \sqrt{36 - 52}}{2}$

(a pair of complex conjugate roots)
\begin{eqnarray*}
{\rm General\  solution:\ } y & = & e_{-3x}(A \cos(2x) +B
\sin(2x))
\\ \frac{dy}{dx} & = & e^{-3x}((2B-2A) \cos (2x) - (2A + 3B)\sin 2x) \\ {\rm Boundary\
conditions:\ } 1 & = & A  \\ 2 & = & 2B - 3A \\ \Rightarrow a & =&
1 {\rm \ and\ } B=\frac{5}{2} \\ {\rm Particular\  solution:\ } y
 & = & e^{-3x}(\cos 2x + \frac{5}{2} \sin 2x)
\end{eqnarray*}

\item[(c)]

Auxiliary equation: $\lambda^2 - 6\lambda + 9  =  0$

Factorises to give: $(\lambda + 3)^2 =  0$

so two repeated real roots $\lambda_1 = \lambda_2 = 3$
\begin{eqnarray*}
{\rm General\  solution:\ } y & = & (A + Bx)e^{3x} \\
\frac{dy}{dx} & = & 3Ae^{3x} + Be^{3x} + 3Bxe^{3x} \\ {\rm
Boundary\ conditions:\ } 2 & = & A  \\ 4 & = & 3A + B \\
\Rightarrow A & = & 2 {\rm \ and\ } B = -2 \\ {\rm Particular\
solution:\ } y & = & 2(1 - x)e^{3x}
\end{eqnarray*}
\end{description}


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