- #1

- 128

- 1

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

Inserting this in the diff. eq. gives:

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

Now, in the other types of diff. eqations I've solved using this method I found a recursive formula for the coefficients, but I don't see how I can do that here? How would I got about solving this problem?

Thanks