Solve $\displaystyle zw'' + w' - w = 0$ where $\displaystyle f(0) = 1$. I'm not exactly sure of how to do it, but I guess it's easier to comment if I show how I'd like to solve it. What I really need help with are the summations, and how to alter the starting integer while rewriting the sums. I tried to get every step on the way here, that's why the post is bloated.

I start out with $\displaystyle \displaystyle w = \sum_{n=0}^\infty a_n z^n$

$\displaystyle \displaystyle w' = \sum_{n=1}^\infty n a_n z^{n-1}$ - I seem to remember adding one, n = 1, when differentiate... Please comment!

$\displaystyle \displaystyle zw'' = z \sum_{n=1}^\infty n(n-1) a_n z^{n-2} = \sum_{n=1}^\infty n(n-1) a_n z^{n-1} $ - This is one thing I'm unsure of. should it be n = 2 in the summation? Please comment!

Back to the equation.

$\displaystyle \displaystyle \sum_{n=1}^\infty n(n-1) a_n z^{n-1} + \sum_{n=1}^\infty n a_n z^{n-1} - \sum_{n=0}^\infty a_n z^n = 0$

$\displaystyle \displaystyle \sum_{n=1}^\infty n^2 a_n z^{n-1} - \sum_{n=0}^\infty a_n z^n = 0$

In the first series use n = k + 1

$\displaystyle \displaystyle \sum_{k = 0}^\infty (k+1)^2 a_{k+1} z^{k} - \sum_{n=0}^\infty a_n z^n = 0$ - Once again I'm unsure of the starting integer of the sum. Is it right to subtract one when I use n = k + 1?

$\displaystyle \displaystyle \sum_{n = 0}^\infty (n+1)^2 a_{n+1} z^{n} - \sum_{n=0}^\infty a_n z^n = 0$

$\displaystyle \displaystyle \sum_{n = 0}^\infty ((n+1)^2 a_{n+1} - a_n) z^{n} = 0$

$\displaystyle \displaystyle (n+1)^2 a_{n+1} = a_n $

$\displaystyle f(0) = 1 \implies a_0 = 1$

$\displaystyle a_n = \frac{1}{(n!)^2}$