Series Solution to Second Order Linear ODE

Hey I had to miss 2 lectures this week due to illness, trying to catch up. Hoping someone can help me with my snag on a problem here, not sure how to deal with 3 different coefficient terms when working with the recursion formula... Thanks! I imagine I'm going wrong somewhere here...

Quote:

$\displaystyle y'' +ty' + (t^2 -1)y=0$ $\displaystyle y(0)=1, y'(0)=2$ has a power series solution about $\displaystyle t=0$ of the form $\displaystyle \displaystyle \sum_{n=0}^\infty C_n t^n

$

So I started by differentiating the power series and substituting into the ODE.

$\displaystyle \displaystyle \sum_{n=2}^\infty n(n-1)C_n t^{n-2} + t \cdot \sum_{n=1}^\infty nC_n t^{n-1} + (t^2 -1)\sum_{n=0}^\infty C_n t^n=0$

Pulling the $\displaystyle t$ and $\displaystyle (t^2 -1)$ terms into the sums.

$\displaystyle \displaystyle \sum_{n=2}^\infty n(n-1)C_n t^{n-2} + \sum_{n=1}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=0}^\infty C_n t^n=0$

So now we need to combine the sums into a single summation so we can equate coefficients for the recursion formula. We need to shift indexes and get the same power of $\displaystyle t$.

$\displaystyle \displaystyle \sum_{n=0}^\infty (n+2)(n+1)C_{n+2} t^{n} + \sum_{n=0}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_{n-2} t^{n} - \sum_{n=0}^\infty C_n t^n=0$

So now we can add coefficients and combine the summations.

$\displaystyle \displaystyle \sum_{n=0}^\infty [(n+2)(n+1)C_{n+2} +nC_n +C_{n-2} -C_n]t^n=0$

Which when we equate coefficients we obtain the following recursion formula;

$\displaystyle \displaystyle C_{n+2}=\frac{(n-1)C_n + C_{n-2}}{(n+2)(n+1)}$

This is where I hit a wall, I've tried plugging increasing values of n in to try to find something, but I am having trouble seeing any sort of pattern or where to go from here... I imagine i've made a mistake somewhere.

Just a push in the right direction would be great, thanks so much!

Kasper