An ODE I was thinking of.

  • #1
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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.
 
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Answers and Replies

  • #2
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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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  • #3
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Thanks.
 
  • #4
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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
 
  • #5
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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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