COuld someone check this out for me and see if If done this right..

I'm worried incase I went wrong somewhere...

$\displaystyle

\begin{gathered}

4\frac{{d^2 y}}

{{dx^2 }} + 8\frac{{dy}}

{{dx}} + 3y = 0 \hfill \\

y = e^{\lambda x} \hfill \\

\hfill \\

4\lambda ^2 + 8\lambda + 3 = 0 \hfill \\

(2\lambda + 1)(2\lambda + 3) = 0 \hfill \\

\lambda = - 1/2 \hfill \\

\lambda = - 3/2 \hfill \\

\end{gathered}

$

So the General solution is..

$\displaystyle

y = Ae^{ - x/2} + Be^{ - 3/2x}

$

So this is the part I'm unsure I'm doing correctly...

I have the two solutions..

$\displaystyle

\begin{gathered}

y(0) = 1 \hfill \\

y(1) = 0 \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

A = 1 - B \hfill \\

0 = - A\frac{1}

{2}(e^{ - 1/2} ) - B\frac{3}

{2}(e^{ - 3/2} ) \hfill \\

\frac{{ - 1}}

{2}(1 - B)(e^{ - 1/2} ) - \frac{3}

{2}B(e^{ - 3/2} ) = 0 \hfill \\

(\frac{1}

{2}(Be^1 - 1)) - B)(e^{ - 3/2} ) = 0 \hfill \\

(\frac{1}

{2}(Be^1 - 1)) - B) = e^{3/2} \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\frac{{Be^1 }}

{2} - \frac{1}

{2} - B = e^{3/2} \hfill \\

Be^1 - 1 - 2B = e^{3/2} \hfill \\

B(e^1 - 2) = e^{3/2} + 1 \hfill \\

B = \frac{{e^{3/2} + 1}}

{{(e^1 - 2)}} \hfill \\

\hfill \\

\end{gathered}

$

if someone could tell me if this is ok..or show me if ive gone wrong..

Even just a tick or a cross would help