# Method of variation of parameters

• May 6th 2009, 11:35 AM
chella182
Method of variation of parameters
The question goes...

Find the general solution of this differential equation, given that a solution of the corresponding homogeneous equation is $y=x$,
$x^2y''+x(x-2)y'-(x-2)y=x^3$

I've checked that $y=x$ is a solution when the equation has a 0 on the RHS, but it's the whole setting $y(x)=u(x)v(x)$ thing I'm stuck with. Obviously that's $y(x)=xv(x)$, but from there I'm stuck. Can anyone help?
• May 6th 2009, 12:17 PM
Jester
Quote:

Originally Posted by chella182
The question goes...

Find the general solution of this differential equation, given that a solution of the corresponding homogeneous equation is $y=x$,
$x^2y''+x(x-2)y'-(x-2)y=x^3$

I've checked that $y=x$ is a solution when the equation has a 0 on the RHS, but it's the whole setting $y(x)=u(x)v(x)$ thing I'm stuck with. Obviously that's $y(x)=xv(x)$, but from there I'm stuck. Can anyone help?

First you want to solve the homogeneous equation (i.e. rhs = 0). This is done using what you said $y = x u$ . Sub. this into your equation
$x^2y''+x(x-2)y'-(x-2)y=0$. You'll get an equation with only u' and u''. Let u' = v which gives a first order you can solve. Then find u.

Then use variation of parameters.
• May 6th 2009, 12:24 PM
chella182
S'alright, I've solved this in the mean time. I didn't understand what I'd written in my notes, but then I remembered what it was. Thanks for the reply, though.
• May 6th 2009, 12:25 PM
HallsofIvy
Quote:

Originally Posted by chella182
The question goes...

Find the general solution of this differential equation, given that a solution of the corresponding homogeneous equation is $y=x$,
$x^2y''+x(x-2)y'-(x-2)y=x^3$

I've checked that $y=x$ is a solution when the equation has a 0 on the RHS, but it's the whole setting $y(x)=u(x)v(x)$ thing I'm stuck with. Obviously that's $y(x)=xv(x)$, but from there I'm stuck. Can anyone help?

Well, if y= xv, then y'= xv'+ v and y"= xv"+ 2v'. What do you get if you put that into your equation?
• May 6th 2009, 12:37 PM
chella182
Like I said, I've sorted it :) cheers, though.