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Non linear second order DE

  1. Oct 9, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must solve [itex]yy''-(y')^2-6xy^2=0[/itex].


    2. Relevant equations
    Not sure.


    3. The attempt at a solution
    I reach something but this doesn't satisfy the original DE...
    Here is my work:
    I divide the DE by [itex]y^2[/itex] to get the new DE [itex]\frac{y''}{y}- \left ( \frac{y'}{y} \right ) ^2-6x=0[/itex]. Now I notice that [itex]\left ( \frac{y'}{y} \right )'=\frac{y''}{y}-1[/itex] so that the DE to solve reduces to [itex]\left ( \frac{y'}{y} \right )'- \left ( \frac{y'}{y} \right ) ^2+1-6x=0[/itex].
    This suggests me to call a new variable [itex]v=\frac{y'}{y}[/itex]. Thus the DE to solve reduces to [itex]v'-v^2+1-6x=0[/itex]. It is separable so I'm extremely lucky. I reach that [itex]\ln y = \int (3x^2+c_1)dx+c_2 \Rightarrow y(x)=e^{x^3+c_1x}+c_2[/itex].
    Hence [itex]y'=(3x^2+c_1)e^{x^3+c_1x}[/itex] and [itex]6xe^{x^3+c_1x}+(3x^2+c_1)^2e^{x^3+c_1x}[/itex]. Plugging these into the original DE doesn't reduces to 0.
    What did I do wrong?
     
  2. jcsd
  3. Oct 9, 2011 #2

    SammyS

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    [itex]\displaystyle \left(\frac{y'}{y} \right )'=\frac{y''}{y}-(y')^2[/itex]

    Added in Edit:

    As pointed out by fluiistic in the following post: The above is wrong. It should be the following.

    [itex]\displaystyle \left(\frac{y'}{y} \right )'=\frac{y''}{y}-\left(\frac{y'}{y}\right)^2[/itex]
     
    Last edited: Oct 9, 2011
  4. Oct 9, 2011 #3

    fluidistic

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    Ah I see! Thanks for pointing this to me. I think you forgot to divide by y^2 the second term.

    Edit: I now reach the result, namely [itex]y(x)=Ae^{3x^2+c_1x}[/itex] (this work). Thank you very much!
     
    Last edited: Oct 9, 2011
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