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Thread: 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.

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

    satisfies

    $\displaystyle

    y(0) = 0
    $

    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.

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

    satisfies

    $\displaystyle

    y(0) = 0
    $

    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 $\displaystyle y' = (2y + 1)x$ right?
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  3. #3
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    so..

    $\displaystyle
    \smallint \frac{{dy}}
    {{2y + 1}} = \smallint xdx
    $


    $\displaystyle
    \frac{1}
    {2}\log (|2y + 1|) = \frac{{x^2 }}
    {\begin{gathered}
    2 \hfill \\
    \hfill \\
    \end{gathered} } + A
    $

    $\displaystyle
    2y = e^{-x^2 } - 1 + A
    $

    $\displaystyle
    y = A + \frac{{e^{-x^2 } - 1}}
    {\begin{gathered}
    2 \hfill \\
    \hfill \\
    \end{gathered} }
    $


    am I close?
    Last edited by dankelly07; Mar 4th 2008 at 08:53 AM.
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  4. #4
    MHF Contributor kalagota's Avatar
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    $\displaystyle \frac{1}{2} \log (2y + 1) = \frac{x^2}{2} + A \Rightarrow \log (2y+1) = x^2 + A$

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

    $\displaystyle 2y+1 = e^{x^2 + A} = Ae^{x^2}$ 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
    Flow Master
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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..
    $\displaystyle 2y + 1 = A e^{x^2}$

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

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

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

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

    You should confirm this solution by substitution into the DE.
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