Thread: ODE Proof: Solution to this will always exist?

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?

But your told that $\displaystyle H(x,y)$ is homogeneous which means it's of the form $\displaystyle H = H\left(\dfrac{y}{x} \right).$

so it will be $\displaystyle H\left(\frac{vx}{x}\right)=H(v)$
and hence it is separable.
Right?

Originally Posted by Liverpool

so it will be $\displaystyle H\left(\frac{vx}{x}\right)=H(v)$
and hence it is separable.
Right?

yup

Thanks.