show that the function $f$ is the solution to the partial differential equation.

Question

Given that $f(x,y)=\phi(xy)$ where $\phi(xy):\Bbb R \rightarrow \Bbb R$ is a differentiable function. Show that $f$ is the solution to the equation: $$\frac{1}{x} \frac{\partial{f}}{\partial{y}} - \frac{1}{y} \frac{\partial {f}}{\partial{x}}=0$$

Answer 1

Let $\phi:\mathbb{R}\to \mathbb{R}$ be a differentiable function, and for any $a\in \mathbb{R}$ let $\phi_a:\mathbb{R}\to \mathbb{R}$ be defined by $\phi_a(t)=\phi(at)$. Then you have $\phi_a'(t)=a\phi'(at)$. By definition of partial derivative, you now have $$ \frac{\partial f}{\partial x} (x,y)=\phi_{y}'(x)=y\phi'(xy),$$ $$ \frac{\partial f}{\partial y} (x,y)=\phi_{x}'(y)=x\phi'(xy).$$ Therefore, the identity follows.

Answer 2

\begin{eqnarray*} \frac{\partial f}{\partial y} &=& \frac{d\phi}{dz} (z=xy) x\\ \frac{\partial f}{\partial x} &=& \frac{d\phi}{dz} (z=xy) y\\ \frac{1}{x} \frac{\partial f}{\partial y} &=& \frac{d\phi}{dz} (z=xy)\\ \frac{1}{y} \frac{\partial f}{\partial x} &=& \frac{d\phi}{dz} (z=xy)\\ \end{eqnarray*}