1. ## Differential equation check?

COuld someone check this out for me and see if If done this right..
I'm worried incase I went wrong somewhere...

$
\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..

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

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

I have the two solutions..

$
\begin{gathered}
y(0) = 1 \hfill \\
y(1) = 0 \hfill \\
\end{gathered}
$

$
\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}
$

$
\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

2. Originally Posted by dankelly07
COuld someone check this out for me and see if If done this right..
I'm worried incase I went wrong somewhere...

$
\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..

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

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

I have the two solutions..

$
\begin{gathered}
y(0) = 1 \hfill \\
y(1) = 0 \hfill \\
\end{gathered}
$

$
\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}
$

$
\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
Looks good to me so far. Of course, you haven't actually written out the solution to the problem yet.