# Reduction of Order

1. Jul 25, 2008

### bishy

I'm making an assumption while trying to solve DEs by reduction of order. I've got a short form equation that I can use to reduce it if and only if I can place the DE into standard form. The standard form with generic notation would be y"+ P(x)y' +G(x)y = 0 where P(x) and G(x) are continuous and on the interval I. I am not sure if I am able to say the following, therefore my question would be is this valid:

Given the DE y" - xy' = 0; y=e^x; I= (0,infinity] is it valid to state that the DE is under standard form where G(x) = 0?

2. Jul 26, 2008

### HallsofIvy

Staff Emeritus
Yes, of course it is! I am wondering what you mean by "y= e^x" immediately after the differential equation. That is clearly not a solution to the equation.

3. Jul 26, 2008

### bishy

ah ok. thats what I thought, I just wasn't sure. As for it not being a solution, I made it up and never bothered checking it. It's not a big a deal, most of the work I have done is confirmed with this.