Find the solution of $$(1-x^2)y'' - 2xy' + 14y = 5x^3.$$
There are no answers ( although I could substitute them), so I'd like to know if my general approach is correct:
1) I would find the solution to the homogeneous equation $$(1-x^2)y'' - 2xy' + 14y = 0$$ which has the solution of the Legendre polynomial with l = $\frac12 (-1 + \sqrt{57}).$ So it is a linear combination of the even and odd solutions.
2) For the inhomogenuous solutions, I took linear combinations of $P_1(x)$ and $P_3(x)$, which gives $$C_1(-2 + 14)x + C_2(-6 + 7)(5x^3 -x) = 5x^3$$.
The coefficient of $C_2$ is then $1$ and the coefficient of $C_1$ is $1/12$.
Is the approach correct?