ODE Proof: Solution to this will always exist?

**Liverpool** · October 8th 2010, 07:54 PM

Problem:
Prove that with the aid of the substitution $y=vx$ , you can solve any
equation of the form:

$y^n f(x) dx + H(x,y) \, (ydx-xdy) = 0$

where H is homogeneous function in x and y.

Solution:
I rearranged the equation to get:

$\left(y^n f(x) + y H(x,y) \right) dx - x H(x,y) dy = 0$

Since $y=vx \implies dy=vdx+xdv$

Using this in the equation:

$\left[ v^nx^n f(x) + vx H(x,vx) \right] dx - x H(x,vx) (vdx+xdv) = 0$

Rearrange :

$v^nx^n f(x)dx - x^2 H(x,vx) dv = 0$

Now, I stopped!

The last equation is almost separable, but that H is bothering me.

Any help?