Solve the nonlinear first order PDE $u_xu_y-u_z^2=cu$

Question

Suppose that $u$ is a function of $(x,y,z)$, then how to solve the equation

$$u_xu_y-u_z^2=cu,$$ where $c$ is a constant.

I want to get the general solutions. But this equation is nonlinear, I don't know how to use the characteristic method.

Answer 1

$$u_xu_y-u_z^2=cu$$ Change of function: $\quad u=\frac{c}{4}v^2$ $$\left(\frac{c}{2}vv_x\right)\left(\frac{c}{2}vv_y\right)-\left(\frac{c}{2}vv_z\right)^2=c\left(\frac{c}{4}v^2\right)$$ $$v_xv_y-v_z^2=1$$ Try separation of variables : $\quad v=f(x)+g(y)+h(z)$ $$f'(x)g'(y)-\left(h'(z)\right)^2=1$$ This is possible only if $\begin{cases} f'(x)=\alpha=\text{constant}\\ g'(y)=\beta=\text{constant}\\ h'(x)=\gamma=\text{constant}\\ \alpha\beta-\gamma^2=1 \end{cases} \quad\implies\quad \begin{cases} f(x)=\alpha\:x+c_1\\ g(y)=\beta\:y+c_2\\ h(x)=\gamma\:z+c_3\\ \alpha\beta-\gamma^2=1 \end{cases}$

$$v(x,y,z)=\alpha\:x+\beta\:y+\gamma\:z+\delta$$

$$\boxed{u(x,y,z)=\frac{c}{4}\left(\alpha\:x+\beta\:y+\gamma\:z+\delta\right)^2}$$ $$\alpha,\beta,\gamma,\delta =\text{constants, with condition }\quad\alpha\beta-\gamma^2=1$$