The differential equation: $xy'' + y =0$ has a regular singular point at $x=0$, using Frobenius' method, we get only one series solution which is:
$$ y_{1} (x) = x - \frac{1}{2} x^2 + \frac{1}{12} x^3 - \frac {1}{144} x^4 + \ldots$$
I'm trying to find the second solution $y_{2} (x)$ using the known equation:
$$y_{2} (x) = y_{1} (x) \int \frac {e^{- \int P(x) dx}}{(y_{1} (x))^{2}} dx$$
Then: $$ y_{2} (x) = y_{1} (x) \int \frac {dx}{ \left(x - \frac{1}{2} x^2 + \frac{1}{12} x^3 - \frac {1}{144} x^4 + \ldots\right)^{2}}$$
My problem is dealing with $(x - \frac{1}{2} x^2 + \frac{1}{12} x^3 - \frac {1}{144} x^4 + ....)^{2}$ , how do I expand this? The expansion in my text gives $ (x^2 - x^3 + \frac{5}{12} x^4 - \frac {7}{72} x^5 +...)$ and then uses long division on it to give $ \frac {1}{x^2} + \frac {1}{x} + \frac {7}{12} + \frac {19}{72} x + ...$ to integrate term by term.
But I don't know how to expand that infinite squared bracket, can someone tell me how?