How to show that a given function is a solution for a partial differential equation. Let $g : \mathbb R \to \mathbb R$ be a differentiable function and let $f(x,y)=x^ng(\frac {y}{x})$, where $n\in\Bbb Z^+$, show that $f$ is a solution to the PDE 
$$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$$
i know what to use, but i have a problem in the simplification, i am stopped , the far i do was:
$x\frac {\partial f}{\partial x}=nx^ng(\frac {y}{x})-x^{n-1}g´(\frac {y}{x})$
$y\frac {\partial f}{\partial y}=\frac {y}{x}g´(\frac {y}{x})$
and thats what i got, besides the sum and a vague simplification but i can´t really approach the answer
 A: Let $h (x, y) = y / x$, so that $f (x, y) = x^n g (y/x)$ can be written in the form $f (x, y) = x^n g ( h (x,y))$. The partial derivatives can be computed using the chain rule, as follows
$\partial_x f (x,y) = n x^{n-1} g (h(x,y)) + x^n g'(h(x,y)) \partial_x h (x,y)$
$\partial_y f (x,y) = x^n g'(h(x,y)) \partial_y h (x,y)$
and, since $\partial_x h (x,y) = -y / x^2$ and $\partial_y h (x,y) = 1 / x$, we obtain
$\partial_x f (x,y) = n x^{n-1} g(y/x) - x^{n-2} y \, g'(y/x)$
$\partial_y f (x,y) = x^{n-1} g'(y/x)$
and, therefore, we have
$x\, \partial_x f (x,y) = n x^n g(y/x) - x^{n-1} y\, g'(y/x)$
$y\, \partial_y f (x,y) = x^{n-1} y\, g'(y/x)$
and, finally, we obtain
$x\, \partial_x f (x,y) + y\, \partial_y f (x,y) = n x^n g(y/x) = n f (x,y)$.
A: I think the comments have pointed out where you went wrong, but I will just give a bit more detail (but as you've listed it as a homework question, I'm not going to do it all).
Basically, you have just calculated the partial derivatives incorrectly. By the product rule, you obtain:
$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(x^ng\left(\frac{y}{x}\right)\right) = \frac{\partial}{\partial x}(x^n)g\left(\frac{y}{x}\right) + x^n\frac{\partial}{\partial x}\left(g\left(\frac{y}{x}\right)\right).$$
Now, you need to use the chain rule to calculate $\frac{\partial}{\partial x}\left(g\left(\frac{y}{x}\right)\right)$. That is:
$$\frac{\partial}{\partial x}\left(g\left(\frac{y}{x}\right)\right) = g'\left(\frac{y}{x}\right)\frac{\partial}{\partial x}\left(\frac{y}{x}\right).$$
To calculate $\frac{\partial f}{\partial y}$ is slightly simpler as you don't need to use the product rule ($x^n$ is not a function of $y$), but you still need to use the chain rule to calculate $\frac{\partial}{\partial y}\left(g\left(\frac{y}{x}\right)\right)$.
Substituting your expressions for $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ into the left hand side of your equation, you should get the right hand side, which shows that $f$ is indeed a solution to the given differential equation.
