# Differential equations

1. Oct 1, 2009

### Axial

So i've just started Differential equations and already im stuck.

I have the Question:

xy' = y² + y It also gives me (u = y/x) and i can see how to get u.

So now i have y' =(y²/x) + u Now im stuck, i'm not really sure of the exact way of going about this.

2. Oct 1, 2009

### Staff: Mentor

This equation is separable, so that you can get dy and the y terms on one side, and dx and the x term on the other. Then integrate, not forgetting the constant of integration.

3. Oct 1, 2009

### HallsofIvy

Staff Emeritus
If you want to treat it as a "homogeneous" equation, which is what the "hint", to take u= y/x implies, then dividing the equation by x gives $y'= y^2/x+ y/x$ as you say and, yes, that is $y'= y^2/x+ u$. But you can also write $y^2/x= (y/x)y= uy$. Your equation becomes y'= uy+ u= u(y+1) which is again separable:
dy/(y+1)= u du.

I tend to agree with Mark44, that, since it is already separable, doing it as dy/(y^2+ y)= dx/x seems more obvious.

4. Oct 1, 2009

### Axial

I would have done it automaticly like that, exept it states * use u=y/x *

This is what the given solution is:

I think i might stick with my answer and see what he says.

5. Oct 1, 2009

### Staff: Mentor

Check your solution by calculating dy/dx. It should be that for your solution xdy/dx = y2 + y. If so, you're in business. If not, you need to go back and see where you went wrong.