Consider the IVP \begin{equation} xu_x-yu_y=xu\\ u(s,s^2)=1 \; \forall s\in \mathbb{R} \end{equation}
I am trying to solve this quasilinear PDE using the method of characteristics, such that I have to solve the following \begin{equation} \frac{dx}{dt}=x,\;\;\frac{dy}{dt}=-y,\;\; \frac{dz}{dt}=xz\\ x(s,s^2)=s,\; y(s,s^2)=s^2,\; z(s,s^2)=1 \end{equation}
I can solve characteristic equation for y, which can be integrated to obtain $y = C_1(s)e^{-t},$ and the initial condition then yields $y = s^2e^{s^2}e^{-t}.$ I am not sure if I parametrized the equations incorrectly from the initial data, but I am having trouble obtaining $x(s,t)$ and $z(s,t)$ from here.
$$\frac{dx}{dt}=x,\;\;\frac{dy}{dt}=-y,\;\; \frac{dz}{dt}=xz\quad\text{ is correct}$$ Or on the Charpit-Lagrange form : $$dt=\frac{dx}{x}=\frac{dy}{-y}=\frac{du}{xu}$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{-y}$: $$xy=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{du}{xu}$ : $$e^{-x}u=c_2$$ General solution of the PDE on the form of implicit equation $c_2=F(c_1)$ : $$e^{-x}u=F(xy)$$ $F$ is an arbitrary function. $$\boxed{u(x,y)=e^xF(xy)}$$ Condition : $u(s,s^2)=1=e^sF(s^3)\quad\implies\quad F(s^3)=e^{-s}$ .
Let $X=s^3$ $$F(X)=e^{-X^{1/3}}$$ Now the function $F(X)$ is determied. We put it into the above general solution where $X=xy$ : $$u(x,y)=e^xe^{-(xy)^{1/3}}$$ $$\boxed{u(x,y)=e^{x-(xy)^{1/3}}}$$