Find the general solution to this Differential Equation.

Question

Given that $z=f(x,y)$ and $a\in \mathbb{R}$ is a constant, I have to solve the following differential equation: $$ \frac{z \,dz+y \,dy}{y^2+z^2}=\frac{dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$ I have not seen anything like this before so any ideas/hints would be much appreciated.

Answer 1

Let rewrite equation: $$\frac{z\,dz+y\,dy}{y^2+z^2}=\frac{dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$ $$\frac{d(y^2+z^2)}{y^2+z^2}=\frac{2dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$ With changing $y^2+z^2=t^2$ and $x-a=s$: $$\frac{2tdt}{t^2}=\frac{2ds}{\sqrt{s^2+t^2}+s}.$$ or $$\frac{\sqrt{s^2+t^2}+s}{t}dt=ds$$ This is a homogeneous equation. Another substitution $s=ut$ gives us $$\frac{dt}{t}=\frac{du}{\sqrt{u^2+1}}$$ and we conclude $\ln(t)=\sinh^{-1}(u)+C$. Now we return main variables.