- #1

MathematicalPhysicist

Gold Member

- 4,459

- 274

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: