If $f(x,y) = g(u)$ and $u=u(x,y)$, find $\frac{g'(u)}{g(u)} (x\frac{\partial u }{\partial x} + y \frac{\partial u}{\partial y})$

Question

If $f(x,y) = g(u)$ and $u=u(x,y)$is a homogenous function of degree $p$. Let $g$ satisfy $g(\lambda u) = \lambda^m g(u)$ and $g(u) \neq 0$. Find $$\frac{g'(u)}{g(u)} (x\frac{\partial u }{\partial x} + y \frac{\partial u}{\partial y})$$ in terms of $m,p$.

I am getting

$$\frac{\partial g} { \partial x} = \frac{\partial g} { \partial u}\frac{\partial u} { \partial x} \\ \frac{\partial g} { \partial y} = \frac{\partial g} { \partial u}\frac{\partial u} { \partial y}$$

Using euler theorem on homogenous function by noting $ g(u(\lambda x,\lambda y)) = g(\lambda^p u(x,y)) = \lambda^{mp}g(u(x,y))$

$$x\frac{\partial g} { \partial x}+y\frac{\partial g} { \partial y} = g'(u) (x\frac{\partial u }{\partial x} + y \frac{\partial u}{\partial y})\\ mp=\frac{g'(u)}{g(u)} (x\frac{\partial u }{\partial x} + y \frac{\partial u}{\partial y})$$

So I am getting answer as $mp$

But given solution is $m$.

I need to know error in this answer, especially the line $ g(u(\lambda x,\lambda y)) = g(\lambda^p u(x,y)) = \lambda^{mp}g(u(x,y))$

Thank you!

Answer 1

You're correct, and the ‘given’ solution is wrong (or else you've transcribed the question wrong). You can check this with a simple example such as $g(u) = u^3$, $u(x,y) = xy$. In this case, $\frac{g′(u)}{g(u)} (x\frac{∂u}{∂x} + y\frac{∂u}{∂y}) = \frac{3u^2}{u^3} (xy + yx) = \frac{6xy}{xy} = 6$, not $3$. Any other example (with $p \ne 1$) will show the same thing.