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**charikaar** $\displaystyle (1 - x^2)y'' - 2xy' + 6y = 0$ given that $\displaystyle u(x)=3x^{2}-1$ is a solution of the equation and satisfies $\displaystyle u(-1)+(1/3)u'(-1)=0$

find another solution $\displaystyle v(x)$ of the equation satisying $\displaystyle v(0)=0$

I tried:

$\displaystyle v(x)=w(x)(3x^{2}-1)$

$\displaystyle v'(x)=(3x^{2}-1)w'+6xw$

$\displaystyle v''=(3x^{2}-1)w''+12xw'+6w$

how do i get to $\displaystyle (3x^{2}-1)(1-x^2)w''=Xw'$ where X is some function. I will be able to go further once i know what X is.

Also when do you use the given boundary conditions?

Many Thanks