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An ODE I was thinking of.

  1. Feb 14, 2013 #1

    MathematicalPhysicist

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    I thought today of the next DE:

    [tex] y''(x) = y(x)e^{y'(x)}[/tex]

    Not sure if it has applications, obviosuly I tried to find a solution via power series around x=0.

    It seems tough to look for a general recurrence equation for the coefficients.
    Here's what I have done so far.

    [tex]y(x)=\sum_{n=0}^{\infty} a_n x^n [/tex]

    [tex]y'(x)=\sum_{n=0}^{\infty} na_n x^{n-1}[/tex]

    [tex]y''(x)=\sum_{n=0}^{\infty} n(n-1)a_n x^{n-2} [/tex]

    equating:

    [tex] \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2} =\sum_{n=0}^{\infty} a_n x^n e^{\sum_{n=0}^{\infty} a_n n x^{n-1}} [/tex]

    [tex]e^{a_1} e^{2a_2 x} e^{3a_3 x^2} \cdots = e^{a_1}[1+2a_1 x + \frac{(2a_1 x)^2}{2!}+\cdots]\cdot [1+3a_3 x^2 +\frac{(3a_3 x^2)^2}{2!}+\cdots]\cdot \cdots [/tex]

    I am not sure if it even converges, is this equation known already, I am quite sure someone already thought of it.

    Thanks in advance.
     
    Last edited: Feb 14, 2013
  2. jcsd
  3. Feb 14, 2013 #2
    Hi !
    The ODE is solvable on the form of the inverse function x(y) as a special function defined by an integral :
     

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  4. Feb 14, 2013 #3

    MathematicalPhysicist

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    Thanks.
     
  5. Feb 15, 2013 #4

    MathematicalPhysicist

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    Well, if I am already in the mood for non-ordinary DEs, I'll make this thread a thread with peculiar DEs I have in my mind.

    Here's another one:

    [tex]y^{(n)}+(y^{(n-1)})^2+(y^{(n-2)})^3+\ldots + (y')^{n+1}+y^{n+2} = 0[/tex]

    Guessing a solution in the form of power series will be hard work (which I don't have time for right now), so does it have an specail function form solution?

    P.S
    n\geq 1
     
  6. Feb 15, 2013 #5

    MathematicalPhysicist

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    Maybe some examples if the general case isn't clear enough.

    for n=1:

    [tex] y'+y^2=0[/tex]

    For n=2:

    [tex] y''+(y')^2+y^3=0[/tex]

    For n=3:

    [tex] y'''+(y'')^2+(y')^3+y^4=0[/tex]

    Etc.
     
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