# [SOLVED]Help with Froebenious method

#### ognik

##### Active member
Hermite's ODE is $y'' - 2xy' + 2\alpha y = 0$

Let $y = \sum_{\lambda = 0}^{\infty} {a}_{\lambda} x^{k+\lambda}, y' = \sum a_{\lambda} (k+\lambda)x^{k+\lambda-1}, y'' = \sum a_\lambda (k+\lambda)(k+\lambda-1)x^{k+\lambda-2}$

I get the indicial eqtn of k(k-1) = 0, therefore k = 0 or 1. Lowest power of x again, lets me choose $a_1=0$

Then using a dummy variable j to make all powers of x equal, then equating coefficients, I get:

$a_{j+2}(k+j+2)(k+j+1) -2a_{j+1}(k+j+1) + 2\alpha a_j = 0$

But the books answer shows me that they found the 2nd term to be $2a_{j}(k+j+1)$ - I can't find what I've done wrong?

#### ognik

##### Active member
Found my mistake, I forgot to multiply the 2nd term by x, so the correct eqtn is:

$a_{j+2}(k+j+2)(k+j+1) -2a_{j}(k+j) + 2\alpha a_j = 0$