Question

Determine $f:R->R$ twice differentiable such that for function  $\varphi$ defined by 

$\varphi(x,y)=f(\frac{x}{y})$ satisfies $\frac{\partial^2\varphi}{\partial x^2}+\frac{\partial^2\varphi}{\partial y^2}=0$

Answer 1

solution

Let $u=\frac{x}{y}=> \varphi(x,y)=f(u)$

$(*)\frac{ \partial\varphi }{\partial x}(x,y)=f'(u)u_x=\frac{1}{y}f'(u)$

$(**)\frac{\partial^2\varphi}{\partial x^2}(x,y)=\frac{1}{y^2}f''(u)$

Then , $\frac{\partial ^2\varphi}{\partial x^2}+\frac{\partial ^2\varphi}{\partial y^2}=\frac{1}{y^2}f''(u)+\frac{2x}{y^3}f'(u)+\frac{x^2}{y^4}f''(u)=0$

$<=>\frac{1}{y^2}[(1+\frac{x^2}{y^2})f''(u)+2\frac{x}{y}f'(u)]=0$

$<=>(1+\frac{x^2}{y^2})f''(u)+2\frac{x}{y}f'(u)=0$

$<=>(1+u^2)f''(u)+2uf'(u)=0$

$<=>[(1+u^2)f'(u)]'=0$

$<=>(1+u^2)f'(u)=k,k\in R$

$<=>\int f'(u) du=k\int \frac{1}{1+u^2}du$

$<=>f(u)=karctan(u)+c,$  $k,c\in R$

Therefore, $\varphi(x,y)=f(\frac{x}{y}) =karctan(u)+c,$  $k,c\in R$

answer 

Therfore , $\varphi(x,y)=f(\frac{x}{y}) =karctan(u)+c,$  $k,c\in R$