Question

to find general solution of
(x^2-1)y"-2xy'+2y=(x^2-1)^2

Answer 1

we have to solve it by method of variation of parameters

Answer 2

Here is what I got from Mathematica
\[
y(x)=\frac{c_1 \sqrt{x^2-1}
(1-x)^{3/2}}{\sqrt{x+1}}-\\
\frac{c_2 x \sqrt{x^2-1}
\sqrt{1-x}}{(x-1) \sqrt{x+1}}+\\
\frac{3 \sqrt{1-x} x^5-3
\sqrt{1-x} x^4-3 \sqrt{1-x} x^3+3 \sqrt{1-x} x^2-6
\sqrt{x+1} \sqrt{1-x^2} x^2+2 \sqrt{x+1} \sqrt{1-x^2}
x-2 \sqrt{x+1} \sqrt{1-x^2} x^4+6 \sqrt{x+1}
\sqrt{1-x^2} x^3}{6 \sqrt{x+1} \sqrt{1-x^2}}
\]

Answer 3

Here is another form
\[
y(x)=\frac{6 c_1 \sqrt{-\left(x^2-1\right)^2} x^2}{6-6
x^2}+ \frac{6 c_2 \sqrt{-\left(x^2-1\right)^2} x}{6-6
x^2}-\\ \frac{12 c_1 \sqrt{-\left(x^2-1\right)^2} x}{6-6
x^2}+ \frac{6 c_1 \sqrt{-\left(x^2-1\right)^2}}{6-6
x^2}- \frac{3 x^2}{6-6 x^2}+\frac{2 x}{6-6
x^2}-\\ \frac{x^6}{6-6 x^2}+\frac{4 x^4}{6-6 x^2}-\frac{2
x^3}{6-6 x^2}
\]

Answer 4

This should be done y a machine and not by hand.