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Thread: Finding a General Solution

  1. #1
    Super Member Deadstar's Avatar
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    Finding a General Solution

    Not sure if this is the right place to post this cos its off my module called Applied Maths but seemed more of a calculus topic...

    $\displaystyle y^{''} + y^{'} + 2y^2 (1 - y) = 0$

    Find the value of the constant a so that $\displaystyle y = (1 + e^{ax})^2$ is a solution of the above equation.

    Now i just plugged the y = ... bit into the equation but it ended as

    ... $\displaystyle = \frac{1}{(1 + e^{ax})^2} (1 - a^2 e^{ax} - \frac{2}{(1 + e^{ax})} + \frac{2a^2 e^{2ax}}{(1 + e^{ax})})$
    and figured something had gone wrong somewhere as i cant solve it from there.

    Any help appreciated
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  2. #2
    Senior Member Peritus's Avatar
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    it's a lot to write...

    $\displaystyle
    \begin{gathered}
    y'' + y' + 2y^2 (1 - y) = 0 \hfill \\
    y_{solution} = \left( {1 + e^{ax} } \right)^2 \hfill \\
    \end{gathered}

    $

    $\displaystyle
    \begin{gathered}
    y' = 2ae^{ax} \left( {1 + e^{ax} } \right) \hfill \\
    y'' = 2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) \hfill \\
    \end{gathered}

    $

    substituting these derivatives into the equation:

    $\displaystyle
    2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) + 2ae^{ax} \left( {1 + e^{ax} } \right) + 2\left( {1 + e^{ax} } \right)^4 \left[ {1 - \left( {1 + e^{ax} } \right)^2 } \right] = 0

    $

    $\displaystyle
    2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) + 2ae^{ax} \left( {1 + e^{ax} } \right) + 2\left( {1 + e^{ax} } \right)^4 \left[ {1 - \left( {1 + e^{ax} } \right)} \right]\left[ {1 + \left( {1 + e^{ax} } \right)} \right] = 0
    $

    $\displaystyle
    2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) + 2ae^{ax} \left( {1 + e^{ax} } \right) + 2\left( {1 + e^{ax} } \right)^4 \left[ { - e^{ax} } \right]\left[ {2 + e^{ax} } \right] = 0

    $

    $\displaystyle
    2e^{ax} \left[ {a^2 + 2a^2 e^{ax} + a + ae^{ax} - \left( {1 + e^{ax} } \right)^4 \left( {2 + e^{ax} } \right)} \right] = 0
    $

    now the outside exponent can nulify the equation only if a->inf which is not interesting.

    $\displaystyle
    a^2 + a + \left( {2a^2 + a} \right)e^{ax} = \left( {1 + e^{ax} } \right)^4 \left( {2 + e^{ax} } \right)

    $

    now because the following equality is supposed to hold for any value of x (y is a solution) then it must hold in particular for x = 0 thus:

    $\displaystyle
    \begin{gathered}
    a^2 + a + 2a^2 + a = 48 \hfill \\
    \hfill \\
    3a^2 + 2a - 48 = 0 \hfill \\
    \hfill \\
    ... \hfill \\
    \end{gathered}

    $
    Last edited by Peritus; Jan 21st 2008 at 02:25 PM.
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  3. #3
    Super Member Deadstar's Avatar
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    Forgot to mention that $\displaystyle y^{''}$ means twice differetiated, etc...
    Is it just substituting in y = ... into the first equation?
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  4. #4
    Super Member Deadstar's Avatar
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    WAIT! its meant to be $\displaystyle y = (1 + e^{ax})^{-1}$ sorry
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  5. #5
    Senior Member Peritus's Avatar
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    Quote Originally Posted by Deadstar View Post
    WAIT! its meant to be $\displaystyle y = (1 + e^{ax})^{-1}$ sorry
    well that's a fatal mistake anyway it doesn't matter what's importent is you understand the main idea:

    because the following equality is supposed to hold for any value of x (y is a solution) then it must hold in particular for x = 0
    just plug x=0 in the expression you've got and solve for a.
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