Question

Solve (1-x^2) y''+2xy'-2y=0, y1=x (one of the solutions)

Answer 1

We introduce an additional variable z such that y=xz. Then

$y' = z + xz'$

$y'' = 2z' + xz''$

$(1-x^2)y'' + 2xy' -2y = (1-x^2)(2z' + xz'') +2x (z+xz') - 2xz = x(1-x^2)z'' + 2z'=0$ $\frac{z''}{z'} = \frac{2}{x(x^2-1)} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{2}{x} = \frac{d}{dx}\ln |\frac{x^2-1}{x^2}| = \frac{d}{dx}\ln |z'|$

from where we have

$|z'| = c|1-x^{-2}|$

if z' is differentiable everywhere besides 0, then $z' = c(1-x^{-2})$

$z = c(x+x^{-1})+c'$

$y=xz=c(x^2+1)+c'x$