I am solving $x^2 y'' + 3xy' + (1 - 2x)y = 0$ using the Method of Frobenius.
The indicial equation of this ODE has the solution $s = -1$ with algebraic multiplicity $2$. It is relatively easy to obtain the first solution $$y = a_0 \sum_{n = 0}^\infty \frac{2^n x^{n - 1}}{n!^2}$$ The method provides a formula for the second solution: $$y = [FIRST SOLUTION]ln|x| + x^{-1} \sum_{n = 1}^\infty \frac{\partial a_n}{\partial s}|_{s = -1} x^n$$ (I'm actually not sure if the $+$ should be a $-$ instead; my book says the former, but the comment below this answer suggests the latter, unless the integer multiple case is different than the algebraic multiplicity case, in this respect)
However, I have seen that Reduction Of Order can also be used to solve this type of problem, so I am trying to do this, starting from the first solution, instead of using the formula to find the second solution.
I plugged $y = v \sum_{n = 0}^\infty \frac{2^n x^{n - 1}}{n!^2}$ into the ODE, where $v$ is a function of $x$, and after some calculus, algebra, and setting to $0$ the part that emerges which is identical to the LHS of the ODE, I obtained $$(\sum_{n = 0}^\infty \frac{2^n x^{n + 1}}{n!^2})v'' + (\sum_{n = 0}^\infty \frac{2^{n + 1}nx^n + 2^n x^n}{n!^2})v' = 0$$ Substituting $w = v'$ and separating variables, this becomes $$w = e^{-\int \frac{\sum_{n = 0}^\infty \frac{2^{n + 1}nx^n + 2^n x^n}{n!^2}}{\sum_{n = 0}^\infty \frac{2^n x^{n + 1}}{n!^2}}dx}$$
However, I am not sure where to go from here. Is there a way to perform this integration, or otherwise solve this ODE using Reduction Of Order without assuming what the form of the solution should be?
