How Does Substitution Work in Solving Homogeneous Differential Equations?

[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.

 

[kdinser]

Just noticed something else odd.

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

 

[marlon]

your book is correct

(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

 

[kdinser]

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

 

[NeutronStar]

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)

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