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Math Help - 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...

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

    Find the value of the constant a so that 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

    ... = \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...

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

    <br />
\begin{gathered}<br />
  y' = 2ae^{ax} \left( {1 + e^{ax} } \right) \hfill \\<br />
  y'' = 2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) \hfill \\ <br />
\end{gathered} <br /> <br />

    substituting these derivatives into the equation:

    <br />
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<br /> <br />

    <br />
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<br />

    <br />
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<br /> <br />

    <br />
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<br />

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

    <br />
a^2  + a + \left( {2a^2  + a} \right)e^{ax}  = \left( {1 + e^{ax} } \right)^4 \left( {2 + e^{ax} } \right)<br /> <br />

    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:

    <br />
\begin{gathered}<br />
  a^2  + a + 2a^2  + a = 48 \hfill \\<br />
   \hfill \\<br />
  3a^2  + 2a - 48 = 0 \hfill \\<br />
   \hfill \\<br />
  ... \hfill \\ <br />
\end{gathered} <br /> <br />
    Last edited by Peritus; January 21st 2008 at 03:25 PM.
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  3. #3
    Super Member Deadstar's Avatar
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    Forgot to mention that 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 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 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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