what is the method for solving a differential equation when a summation is involved from the start ?
ex. what method is required to find a particular solution to
$$(1-x^2)y''-2xy'=\sum_{n=1}^{\infty}\tfrac{P_n(x)}{2^n}$$
I should point out also that
$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}\tfrac{P_n(x)}{2^n}$.
Edit: $|x| \leq 1$
My attempt :
$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}\tfrac{P_n(x)}{2^n} \Rightarrow \sum_{n=0}^{\infty}\tfrac{2^{n+1}}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}P_n(x)$. ( I don't know if you can do this though.....)
then we assume that $y=\sum_{n=0}^{\infty}a_nP_n(x)=\sum \tfrac{a_n2^{n+1}}{(5-4x)^{1/2}}$
$y'=\sum \tfrac{a_n2^{n+2}}{(5-4x)^{3/2}}$
$y''=\sum \tfrac{3a_n2^{n+3}}{(5-4x)^{5/2}}$
Is this anyway right ?
It doesn't matter what function you have on the right hand side. Let's call it:
$$f(x)=\sum_{n=1}^{\infty}\frac{P_n(x)}{2^n}$$
As @Nosrati correctly noted, you have:
$$(1-x^2)y''-2xy'=((1-x^2)y')'$$
Which immediately allows us to integrate both sides:
$$(1-x^2)y'=\int_a^x f(t) dt$$
Where $a$ is an arbitrary constant, which depends on the intitial conditions. I have written the integral this way to keep track of the different integration variables we are going to have.
Now we assume $x \neq \pm 1$ and find:
$$y(x)=\int_b^x \frac{1}{1-u^2} \int_a^u f(t) dt ~du$$
Where the second constant $b$ also depends on the initial conditions.
Now whatever the definition of $f(x)$ (a series or a closed form), you just need to evaluate the integrals and apply the initial conditions correctly.