Given that :
$$((1-x^2)y')'=\tfrac{2}{(5-4x)^{1/2}}-1$$
and that
$$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}\tfrac{P_n(x)}{2^n}$$
we have that
$$((1-x^2)y')'=\sum_{n=1}\tfrac{P_n(x)}{2^n}$$
We want to find a particular solution for y.
well first we not that by Rodrigues's formula $P_n(x)=\tfrac{1}{2^nn!}\tfrac{d^n}{dx^n}(x^2-1)^n$ So now we have that
$$(1-x^2)y'=\sum_{n=1}\tfrac{1}{4^nn!}\int_a^u\tfrac{d^n}{dx^n}(t^2-1)^ndt$$
and that
$$y=\sum_{n=1}\tfrac{1}{4^nn!}\int^x_b\tfrac{1}{1-u^2}\int_a^u \tfrac{d^n}{dt^n}(t^2-1)^ndtdu$$
$$\Rightarrow y = \sum_{n=1}\tfrac{1}{4^nn!}\int_1^x\tfrac{1}{1-u^2}[\tfrac{d^{n-1}}{du^{n-1}}(u^2-1)^n]du$$
At this point I'd like to note something for clarification: I believe that given that a and b are determined by initial values and because we were not given initial values at the beginning of the question that we can set them to be whatever we like therefore i have set a=b=1 to tidy up the equation.
My question :
i) Is this at all the correct way to even find the particular ssolution for y?
ii) if it is then how does one solve the integral on the rhs ? I've tried substitution and integration by parts but nothing seems to work.
iii) if I'm going about this the completely wrong way, or there is an easier way to do it, would anyone be kind enough to suggest an alternative method ?