# Thread: Differential equation check?

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...

$\displaystyle \begin{gathered} 4\frac{{d^2 y}} {{dx^2 }} + 8\frac{{dy}} {{dx}} + 3y = 0 \hfill \\ y = e^{\lambda x} \hfill \\ \hfill \\ 4\lambda ^2 + 8\lambda + 3 = 0 \hfill \\ (2\lambda + 1)(2\lambda + 3) = 0 \hfill \\ \lambda = - 1/2 \hfill \\ \lambda = - 3/2 \hfill \\ \end{gathered}$

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

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...

$\displaystyle \begin{gathered} 4\frac{{d^2 y}} {{dx^2 }} + 8\frac{{dy}} {{dx}} + 3y = 0 \hfill \\ y = e^{\lambda x} \hfill \\ \hfill \\ 4\lambda ^2 + 8\lambda + 3 = 0 \hfill \\ (2\lambda + 1)(2\lambda + 3) = 0 \hfill \\ \lambda = - 1/2 \hfill \\ \lambda = - 3/2 \hfill \\ \end{gathered}$

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