A question from my calculus exam:
Let $u=xyf(\frac{x+y}{xy})$ where $f:\mathbb R\to\mathbb R$ is differentiable. Show that $u$ satisfies the partial differential equation:
$$x^2u_x-y^2u_y-g(x,y)u=0$$
Also find $g(x,y).$
I am not getting the expression.Please help...
$$u(x,y)=xyf(\frac{x+y}{xy})$$ Differentiating wrt $x$ you get $$u_x=yf(\frac{x+y}{xy})+xyf'(\frac{x+y}{xy})\cdot(-\frac{1}{x^2})$$ Differentiating wrt $y$ you get $$u_y=xf(\frac{x+y}{xy})+xyf'(\frac{x+y}{xy})\cdot(-\frac{1}{y^2})$$ So $$x^2\cdot u_x-y^2\cdot u_y+g(x,y)u=0 \Leftrightarrow x^2(yf(\frac{x+y}{xy})+xyf'(\frac{x+y}{xy})\cdot(-\frac{1}{x^2}))-y^2(xf(\frac{x+y}{xy})+xyf'(\frac{x+y}{xy})\cdot(-\frac{1}{y^2}))+g(x,y)u=0 \Leftrightarrow x^2yf(\frac{x+y}{xy})-y^2xf(\frac{x+y}{xy})+g(x,y)xyf(\frac{x+y}{xy})=0\Leftrightarrow xyf(\frac{x+y}{xy})(x-y+g(x,y))=0$$ Assuming $f$ is not identically zero then $$g(x,y)=y-x$$