# Non-homogeneous differential equation

1. Sep 27, 2007

### kasse

How can I solve (x^2)y'' - 2xy + 2y = (x^3)sinx ?

I've used Euler-Caucy to find the homogeneous eq. (C1)e^x + (C2)e^2x. Then I've calculated the wronski (either e^3x or -e^3x depending on which function is y1 and y2). The rest involves solving the integral of xsin(x)/e^x and xsin(x)/e^2x, whics seems quite difficult. Am I on the right track?

2. Sep 27, 2007

### HallsofIvy

You've done that wrong. If y= ex, then y'= y"= ex. Putting those into the homogeneous equation, $x^2e^x- 2xe^2+ 2e^x= (x^2- 2x+2)e^x$ NOT 0. The same is true for e2x: it is clearly NOT a solution. (I'm assuming that "2xy" was supposed to be 2xy'.) I thought perhaps you had confused this with a "constant coefficients" equation but ex and e2x do not satisfy y"- 2y'+ 2y= 0 either, the satisfy y"- 3y'+2y=0. This is an "Euler type" or "equi-potential" equation. You can convert it into a constant coefficients equation, in t, by making the change of variable t= ln(x). Hmm. It converts to y"- 3y'+ 2y= 0 which DOES have et and e2t as solutions! It looks like eln(x)= x and e2ln(x)= x2 are solutions to your homogeneous equation.

Redo the Wronskian, using y1= x and y2= x2.

3. Sep 27, 2007

### kasse

Ah, of course, Euler-Cauchy. Thanks, now I got the right answer.