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Math Help - differential eq solution

  1. #1
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    differential eq solution

    Just wondering if anyone could have a look through nd see if I've done this right? When it comes to the general solution & conditon I'm not sure if I've done this right.

    <br />
\begin{gathered}<br />
  y' - 2xy = x \hfill \\<br />
  y' = 2xy + x \hfill \\<br />
  \smallint \frac{{dy}}<br />
{y} = \smallint 3x \hfill \\<br />
  \log (y) = \frac{3}<br />
{2}x^2  \hfill \\<br />
  y(x) = A\exp \{ \frac{3}<br />
{2}x^2 \}  \hfill \\<br />
\end{gathered} <br />

    satisfies

    <br /> <br />
y(0) = 0<br />

    is the answer y(x)=A??
    I think I might have gone wrong somewhere
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by dankelly07 View Post
    Just wondering if anyone could have a look through nd see if I've done this right? When it comes to the general solution & conditon I'm not sure if I've done this right.

    <br />
\begin{gathered}<br />
  y' - 2xy = x \hfill \\<br />
  y' = 2xy + x \hfill \\<br />
  \smallint \frac{{dy}}<br />
{y} = \smallint 3x \hfill \\<br />
  \log (y) = \frac{3}<br />
{2}x^2  \hfill \\<br />
  y(x) = A\exp \{ \frac{3}<br />
{2}x^2 \}  \hfill \\<br />
\end{gathered} <br />

    satisfies

    <br /> <br />
y(0) = 0<br />

    is the answer y(x)=A??
    I think I might have gone wrong somewhere
    yes.. on the third line
    check your factoring..
    it should be y' = (2y + 1)x right?
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  3. #3
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    so..

    <br />
\smallint \frac{{dy}}<br />
{{2y + 1}} = \smallint xdx<br />


    <br />
\frac{1}<br />
{2}\log (|2y + 1|) = \frac{{x^2 }}<br />
{\begin{gathered}<br />
  2 \hfill \\<br />
   \hfill \\ <br />
\end{gathered} } + A<br />

    <br />
2y = e^{-x^2 }  - 1 + A<br />

    <br />
y = A + \frac{{e^{-x^2 }  - 1}}<br />
{\begin{gathered}<br />
  2 \hfill \\<br />
   \hfill \\ <br />
\end{gathered} }<br />


    am I close?
    Last edited by dankelly07; March 4th 2008 at 09:53 AM.
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  4. #4
    MHF Contributor kalagota's Avatar
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    \frac{1}{2} \log (2y + 1) = \frac{x^2}{2} + A \Rightarrow \log (2y+1) = x^2 + A

    if you take the exponential of both sides, it should turn out to be like this..

    2y+1 = e^{x^2 + A} = Ae^{x^2} and not the one have written.. continue now..
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  5. #5
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    thanks for your help..

    so

    Ae^(x^2)+1/2

    with condition A=1??

    I want to be sure I can do this..
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  6. #6
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    mr fantastic's Avatar
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    Quote Originally Posted by dankelly07 View Post
    thanks for your help..

    so

    Ae^(x^2)+1/2 Mr F says: It's - not + 1/2.

    with condition A=1??

    I want to be sure I can do this..
    2y + 1 = A e^{x^2}

    \Rightarrow y = Ae^{x^2} {\color{red} - }\frac{1}{2}.

    Substitute the given boundary condition y(0) = 0, that is, y = 0 when x = 0:

    0 = A - \frac{1}{2} \Rightarrow A = \frac{1}{2}.

    Therefore y = \frac{1}{2} e^{x^2} - \frac{1}{2}.

    You should confirm this solution by substitution into the DE.
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