Problem:
Prove that with the aid of the substitution $\displaystyle y=vx$, you can solve any
equation of the form:
$\displaystyle y^n f(x) dx + H(x,y) \, (ydx-xdy) = 0$
where H is homogeneous function in x and y.
Solution:
I rearranged the equation to get:
$\displaystyle \left(y^n f(x) + y H(x,y) \right) dx - x H(x,y) dy = 0$
Since $\displaystyle y=vx \implies dy=vdx+xdv$
Using this in the equation:
$\displaystyle \left[ v^nx^n f(x) + vx H(x,vx) \right] dx - x H(x,vx) (vdx+xdv) = 0 $
Rearrange :
$\displaystyle v^nx^n f(x)dx - x^2 H(x,vx) dv = 0$
Now, I stopped!
The last equation is almost separable, but that H is bothering me.
Any help?