1. ODE Proof #2

Problem:
If F is homogeneous of degree k in x and y, F can be written in the form:

$F=x^k g\left(\frac{y}{x}\right)$

Use that to prove the following equation for F:

$xF_x+yF_y=kF$

Solution:
I found that:

$xF_x=kx^kg\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)$

and

$yF_y=x^{k-1} y g_y\left(\frac{y}{x}\right)$

$xF_x+yF_y=kx^k g\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)+x^{k-1} y g_y\left(\frac{y}{x}\right)$

Now, I stopped!

any help?

2. Originally Posted by Liverpool
Problem:
If F is homogeneous of degree k in x and y, F can be written in the form:

$F=x^k g\left(\frac{y}{x}\right)$

Use that to prove the following equation for F:

$xF_x+yF_y=kF$

Solution:
I found that:

$xF_x=kx^kg\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)$

and

$yF_y=x^{k-1} y g_y\left(\frac{y}{x}\right)$

$xF_x+yF_y=kx^k g\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)+x^{k-1} y g_y\left(\frac{y}{x}\right)$

Now, I stopped!

any help?
In order to finish this, you would have to assume that $g_x(y/x)=g_y(y/x)$.

So maybe you're supposed to prove that $xF_x+yF_y=kF \iff g_x(y/x)=g_y(y/x)$?

3. Nope.
The question is as i write it.

4. Chris is right! Another way to look at it is since $g(\cdot)$ is a function of a single argument then

$\dfrac{\partial}{\partial x} \;g \left(\dfrac{y}{x}\right) = - \dfrac{y}{x^2} g ' \left(\dfrac{y}{x}\right)$ and $\dfrac{\partial}{\partial y} \;g \left(\dfrac{y}{x}\right) = \dfrac{1}{x} g ' \left(\dfrac{y}{x}\right)$.

5. Maybe there is something missing in the problem.