I am solving this differential equation using Mathematica $$x ~y''(x)+\alpha ~ y'(x)+\frac{1-3\mu (x)}{(1-x) x}=0$$ where $\mu(x)$ is a polynomial: $$\mu (x)=a+b\left(\frac{1}{1-x}\right)+c\left(\frac{1}{1-x}\right)^2+d\left(\frac{1}{1-x}\right)^3+e\left(\frac{1}{1-x}\right)^4+f\left(\frac{1}{1-x}\right)^5$$
The problem comes here. If I solve it directly (just replacing the expression of $\mu(x)$ I get a specific solution (really long and complex). Then I though about splitting the polynomial , that is: $$x ~y''(x)+\alpha ~ y'(x)+\frac{1-3(a+b\left(\frac{1}{1-x}\right))}{(1-x) x}+\frac{-3c(\frac{1}{1-x})^2}{(1-x) x}+\frac{-3d\left(\frac{1}{1-x}\right)^3}{(1-x) x}+\frac{-3e\left(\frac{1}{1-x}\right)^4}{(1-x) x}+\frac{-3f\left(\frac{1}{1-x}\right)^5}{(1-x) x}=0$$ I didn't think a lot if this was going to work and I solved the following: $$x ~y''(x)+\alpha ~ y'(x)+\frac{1-3(a+b\left(\frac{1}{1-x}\right))}{(1-x) x}=0$$ $$x ~y''(x)+\alpha ~ y'(x)+\frac{-3c\left(\frac{1}{1-x}\right)^2}{(1-x) x}=0$$ $$x ~y''(x)+\alpha ~ y'(x)+\frac{-3d\left(\frac{1}{1-x}\right)^3}{(1-x) x}=0$$ $$x ~y''(x)+\alpha ~ y'(x)+\frac{-3e\left(\frac{1}{1-x}\right)^4}{(1-x) x}=0$$ $$x ~y''(x)+\alpha ~ y'(x)+\frac{-3f\left(\frac{1}{1-x}\right)^5}{(1-x) x}=0$$ wondering that the sum of my solutions would be the same as the total solution computed at the beginning. Now I see this makes no sense (right?). They don't have to be the same (a funny thing is that if a plot both solutions it seems like they have the same shape but are just shifted in the Y axis). My question is, Is there any way of taking advantage of this splitting in order to find a simpler result?
Thank you!
(I editted the formula of $\mu(x)$, it was not correct)
Given a non homogeneous linear differential equation $$ a_2(x)y''+a_1(x)y'+a_0(x)y=b(x) $$ then the general integral is a linear combination of two independent solution of the associated homogeneous equation, say $y_1(x)$ and $y_2(x)$, plus a particular solution of the non-homogeneous equation, say $y_p(x)$: $$ y(x;c_1,c_2)=c_1y_1(x)+c_2y_2(x)+y_p(x) $$ When the term $b(x)$ is a sum of two (or more) terms, say $b(x)=b_1(x)+b_2(x)$, then it is easy to see that $y_p(x)=y_{p,1}(x)+y_{p,2}(x)$ is a solution of the non-homogeneous equation, where $y_{p,1}(x)$ is a particular solution taking into account only $b_1(x)$, and similarly for $y_{p,2}(x)$.
So yes, it is possible to split the term $b(x)$ in a sum, solve the reduced equations then sum the results, but the initial conditions, required to determine the constant $c_1$, $c_2$ must be applied to the full solution.