Prove that a function is k homogenous

Question

Question: Given that $F: \Bbb R^2 \to \Bbb R $ is differentiable and $\forall (x,y) \in \Bbb R^2 :xf_x(x,y)+yf_y(x,y)=kf(x,y)$ Prove that f is homogenous of k'th degree for some k>0.

Thoughts: We've managed to show that using a composition with the function $h(t)=(tx,ty) $s.t $g=f \circ h$ then $g(t)=\frac tk g'(t)$

This is Euler's theorem. I think posting it as an answer is not good because you can find it easily. :-) — Mikasa, Jan 08 '14 at 14:46
We read this now. The problem is we haven't studied differential equations yet. Is there any way to prove it without using these theorems? — jreing, Jan 08 '14 at 15:46
See this , but I don't think we can overcome the problem without using an ODE — Mikasa, Jan 08 '14 at 16:44

score 0 · Accepted Answer · answered Jan 08 '14 at 19:08

Oh, you are really close.

Just integrate both parts, you get that $$ \int \frac{g'(t)}{g(t)}\, \mathrm{d}t = \int \frac{k}{t}\, \mathrm{d}t$$

$$ ln(g(t)) = ln(t^{k}) $$

$$ f(tx,ty) = t^{k} $$

for $t=1$ we get $ f(x,y) = 1 $

therefore $$ f(tx,ty) = t^{k} f(x,y) $$

Prove that a function is k homogenous

1 Answers1