Guys I really don't know what to do, I apply the method of characteristics, but the system of equations it gave me looks unfamiliar and complicated. Any hint on how to approach this?
Find a solution of the first order quasi-linear partial differential equation:
$$z(x +z)\frac{\partial z}{\partial x} + y(y +z)\frac{\partial z}{\partial y} =0$$ such that it passes through points of parabolic cylinder $z = \sqrt y$ and $x=1$.
Thanks.
On
The given condition implies that the curve $\Gamma = \{ (1,s^2,s) \in \mathbb{R}^3 : s \ge 0 \} $ must be contained in the graph of our solution. We apply the characteristics method and transform the initial PDE into the ODE system: \begin{cases} \dot x = z(x+z) \\ \dot y=y(y+z) \\ \dot z=0 \end{cases}
The last equation immediately gives $z(t)=c_1=c_1(s)$, hence we are left to solve: \begin{cases} \dot x =c_1 x + c_1^2 \\ \dot y = y^2 + c_1 y \end{cases} The first equation is a classic first order linear non-homogeneous, while the second is first-order separable. With standard ODEs integration we get (hope I didn't mess up with those constants): \begin{cases} x(t,s)=c_2(s) e^{c_1(s)t}-c_1(s) \\ y(t,s)=-\frac{c_1(s)c_3(s)e^{c_1(s)t}}{c_3(s)e^{c_1(s)t}-1} \\ z(t,s)=c_1(s) \end{cases}
Now it suffices to impose the initial condition (i.e, the curve $\Gamma$) to determine$c_1,c_2,c_3$ and finally, if possible, eliminate the parameters $t,s$ to get a solution $z(x,y)$.
The result below is an implicit equation : $$z^3+x\,z^2+(x-2)\,y\,z-y=0$$ The solution could be expressed on explicit form in solving the cubic equation for $z$.