# Solving a homogeneous diff eq

1. Jan 25, 2005

### kdinser

The book only has one example of this and it's really confusing me.

$$(x^2+y^2)dx+(x^2-xy)dy=0$$

I can see that it's homogeneous of degree 2

They then let $$y = ux$$
From there they state that $$dy = udx+xdu$$ (I'm not sure where this is coming from, but can just accept it on faith if I have to)

I'm fine with making the subs.

$$(x^2+u^2x^2)dx+(x^2-ux^2)(u dx+x du)=0$$

This is the part that really screws me up.

$$x^2(1+u)dx+x^3(1-u)du=0$$

Where did the $$x^3$$ come from? All I see is $$x^2$$. Or I guess I should ask, how did $$udx+xdu$$ become just $$xdu$$? That would explain the $$x^3$$ they have.

2. Jan 25, 2005

### kdinser

Just noticed something else odd.

How did they turn $$(x^2+x^2u^2)$$ into $$x^2(1+u)$$ what happend to the other u? Shouldn't it be $$x^2(1+u^2)$$?

3. Jan 25, 2005

### marlon

$$(x^2+u^2x^2)dx+(x^2-ux^2)(u dx+x du)=0$$

$$(x^2+u^2x^2)dx + ux^2dx -u^2x^2dx + x^3du -ux^3du=0$$
just calculate the above formula...you must have mage an error there

how do you do this ???

marlon

Last edited: Jan 25, 2005
4. Jan 25, 2005

### kdinser

Thanks, I got it now, I just wasn't carrying things far enough.

5. Jan 25, 2005

### NeutronStar

It's just the product rule of differentiation:

$$y = ux$$

$$\frac{d}{du}\left[ y\right] =\frac{d}{du}\left[ ux\right]$$

$$\frac{dy}{du} = u\frac{dx}{du}+x\frac{du}{du}$$

$$dy = udx+xdu$$