I've got this for homework:
Find general solution for this first order PDE
$\displaystyle (\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2= \frac{k}{r} - h$
where $k$ and $h$ are constant real numbers and $ r=\sqrt{x^2+y^2}$ .
Since $\displaystyle (\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2 = \vert \nabla u\vert^2$ , it is clear, from the upper equation, that $\lvert \nabla u \lvert $ depends only on $r$ which gave me an idea to look for a radial solution $u(x,y)=g(r)$. My question would be: is this only solution and how can I know if it is?
Hint:
$\left(\dfrac{\partial u}{\partial x}\right)^2+\left(\dfrac{\partial u}{\partial y}\right)^2=\dfrac{k}{\sqrt{x^2+y^2}}-h$
$\left(\dfrac{\partial u}{\partial x}\right)^2=\dfrac{k}{\sqrt{x^2+y^2}}-h-\left(\dfrac{\partial u}{\partial y}\right)^2$
$\dfrac{\partial u}{\partial x}=\pm\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-\left(\dfrac{\partial u}{\partial y}\right)^2}$
$\dfrac{\partial^2u}{\partial x\partial y}=\pm\dfrac{-\dfrac{ky}{(x^2+y^2)^\frac{3}{2}}-2\dfrac{\partial u}{\partial y}\dfrac{\partial^2u}{\partial y^2}}{2\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-\left(\dfrac{\partial u}{\partial y}\right)^2}}$
Let $v=\dfrac{\partial u}{\partial y}$ ,
Then $\dfrac{\partial v}{\partial x}=\pm\dfrac{-\dfrac{ky}{(x^2+y^2)^\frac{3}{2}}-2v\dfrac{\partial v}{\partial y}}{2\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-v^2}}$
$\dfrac{\partial v}{\partial x}\pm\dfrac{v}{\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-v^2}}\dfrac{\partial v}{\partial y}=\mp\dfrac{ky}{2(x^2+y^2)^\frac{3}{2}\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-v^2}}$