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Math Help - Differential equation check?

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    76

    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...

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

    So the General solution is..

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

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

    I have the two solutions..

    <br />
\begin{gathered}<br />
  y(0) = 1 \hfill \\<br />
  y(1) = 0 \hfill \\ <br />
\end{gathered} <br />



    <br />
\begin{gathered}<br />
  A = 1 - B \hfill \\<br />
  0 =  - A\frac{1}<br />
{2}(e^{ - 1/2} ) - B\frac{3}<br />
{2}(e^{ - 3/2} ) \hfill \\<br />
  \frac{{ - 1}}<br />
{2}(1 - B)(e^{ - 1/2} ) - \frac{3}<br />
{2}B(e^{ - 3/2} ) = 0 \hfill \\<br />
  (\frac{1}<br />
{2}(Be^1  - 1)) - B)(e^{ - 3/2} ) = 0 \hfill \\<br />
  (\frac{1}<br />
{2}(Be^1  - 1)) - B) = e^{3/2}  \hfill \\ <br />
\end{gathered} <br />


    <br />
\begin{gathered}<br />
  \frac{{Be^1 }}<br />
{2} - \frac{1}<br />
{2} - B = e^{3/2}  \hfill \\<br />
  Be^1  - 1 - 2B = e^{3/2}  \hfill \\<br />
  B(e^1  - 2) = e^{3/2}  + 1 \hfill \\<br />
  B = \frac{{e^{3/2}  + 1}}<br />
{{(e^1  - 2)}} \hfill \\<br />
   \hfill \\ <br />
\end{gathered} <br />

    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
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  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,720
    Thanks
    1472
    Quote Originally Posted by dankelly07 View Post
    COuld someone check this out for me and see if If done this right..
    I'm worried incase I went wrong somewhere...

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

    So the General solution is..

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

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

    I have the two solutions..

    <br />
\begin{gathered}<br />
  y(0) = 1 \hfill \\<br />
  y(1) = 0 \hfill \\ <br />
\end{gathered} <br />



    <br />
\begin{gathered}<br />
  A = 1 - B \hfill \\<br />
  0 =  - A\frac{1}<br />
{2}(e^{ - 1/2} ) - B\frac{3}<br />
{2}(e^{ - 3/2} ) \hfill \\<br />
  \frac{{ - 1}}<br />
{2}(1 - B)(e^{ - 1/2} ) - \frac{3}<br />
{2}B(e^{ - 3/2} ) = 0 \hfill \\<br />
  (\frac{1}<br />
{2}(Be^1  - 1)) - B)(e^{ - 3/2} ) = 0 \hfill \\<br />
  (\frac{1}<br />
{2}(Be^1  - 1)) - B) = e^{3/2}  \hfill \\ <br />
\end{gathered} <br />


    <br />
\begin{gathered}<br />
  \frac{{Be^1 }}<br />
{2} - \frac{1}<br />
{2} - B = e^{3/2}  \hfill \\<br />
  Be^1  - 1 - 2B = e^{3/2}  \hfill \\<br />
  B(e^1  - 2) = e^{3/2}  + 1 \hfill \\<br />
  B = \frac{{e^{3/2}  + 1}}<br />
{{(e^1  - 2)}} \hfill \\<br />
   \hfill \\ <br />
\end{gathered} <br />

    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.
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