solutions of ODE

**charikaar** · December 6th 2009, 03:38 AM

$(1 - x^2)y'' - 2xy' + 6y = 0$ 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

**Danny** · December 6th 2009, 07:05 AM

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.

**charikaar** · December 6th 2009, 07:46 AM

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 $X=14x-18x^3$

**Danny** · December 6th 2009, 08:23 AM

Originally Posted by charikaar

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 $X=14x-18x^3$

Yep, that's what I got except for a negative, i.e.

$(3x^2-1)(1-x^2)w'' + (14x-18x^3)w'=0$ .

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