- #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.
[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: