1. ## solutions of ODE

$(1 - x^2)y'' - 2xy' + 6y = 0$ given that $u(x)=3x^{2}-1$ is a solution of the equation and satisfies $u(-1)+(1/3)u'(-1)=0$
find another solution $v(x)$ of the equation satisying $v(0)=0$
I tried:
$v(x)=w(x)(3x^{2}-1)$
$v'(x)=(3x^{2}-1)w'+6xw$
$v''=(3x^{2}-1)w''+12xw'+6w$
how do i get to $(3x^{2}-1)(1-x^2)w''=Xw'$ 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

2. Originally Posted by charikaar
$(1 - x^2)y'' - 2xy' + 6y = 0$ given that $u(x)=3x^{2}-1$ is a solution of the equation and satisfies $u(-1)+(1/3)u'(-1)=0$
find another solution $v(x)$ of the equation satisying $v(0)=0$
I tried:
$v(x)=w(x)(3x^{2}-1)$
$v'(x)=(3x^{2}-1)w'+6xw$
$v''=(3x^{2}-1)w''+12xw'+6w$
how do i get to $(3x^{2}-1)(1-x^2)w''=Xw'$ 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
If you substitute your variable and its derivatives v, v' and v'' into the original and simplify you'll get your desired form. Once you reduce the second order ODE by using $W = w'$, you can separate and integrate, then integrate again. At the end use you initial condition (or boundary conditon) to find your constants of integration.

3. Thank you for your help.
Do you mean I should plug

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

I get $X=14x-18x^3$

4. Originally Posted by charikaar
$y=v(x)=w(x)(3x^{2}-1)$
$y'=v'(x)=(3x^{2}-1)w'+6xw$
$y''=v''=(3x^{2}-1)w''+12xw'+6w$
I get $X=14x-18x^3$
$(3x^2-1)(1-x^2)w'' + (14x-18x^3)w'=0$.