# Modeling with DE's

1. Jul 27, 2008

### franky2727

I have a question of show that y=x is one solution of the equation (1+x^2)y''-2xy'+2y=0) this is a question that i need to know for a coming exam. i understand and can do everything until the question gets to v''+(2/x(1+x^2))v'=0

then the question skips to IF =e^integral of (2/x(1+x^2))dx. what does this mean and where does it come from

2. Jul 27, 2008

### Hootenanny

Staff Emeritus
IF is known as the integrating factor as it one technique used to solve ODE's.

3. Jul 27, 2008

### franky2727

ok thanks and why is the integrating factor that part of the equation?

4. Jul 27, 2008

### Hootenanny

Staff Emeritus

Do you follow, or does it need more explanation?

5. Jul 27, 2008

### HallsofIvy

Staff Emeritus
If the problem is, as you say, just to show that y= x is one solution then you don't need to do all that. If y= x, then y'= 1 and y"= 0 so the equation becomes
(1+ x2)(0)- 2x(1)+ 2(x)= -2x+ 2x= 0. Q.E.D.

However, I suspect that the problem asks you to (1) show that y= x is a solution to the equation and then (2) find another, independent, solution.

Since y= x is a solution, we can look for another solution of the form y= xv(x) where v(x) is an unknown function of x. Now y'= xv'+ v and y"= xv"+ 2v'. Putting those into the equation, (1+ x^2)(xv"+ 2v')- 2x(xv'+ v)+ 2xv= (x+ x^3)y"+ (2+ 2x^2- 2x^2)v' - 2xv+ 2xv= (x+ x^3)v"+ 2v'= 0. While that is a second order equation, v itself does not appear in the equation (because y= x satisfies the equation) and, letting u= v', we have a first order equation, (x+ x^3)u'+ 2u= 0. Now you can use the "integrating factor" site Hootenanny gave to solve that first order equation for u, then integrate to find v and, finally, find y= xv.