Stuck solving this PDE problem I need to solve $$uu_x+yu_y=x, u(x,1)=2x$$ i tried with characteristics method but got stuck solving the ODEs $$\frac{dx}{dt}=u ~,~\frac{dy}{dt}=y~,~\frac{du}{dt}=x$$ with initial values $x(0)=s,y(0)=1,u(0)=2s$. From the second one i get $y=e^t$, but how to solve the first two ODEs? I tried $$\frac{dx}{du}=\frac{u}{x} \implies x^2-u^2=k_1$$ where $k_1=-3s^2$ which gives $u=\sqrt{x^2+3s^2}$. Similarly $$\frac{dy}{y}=\frac{dx}{\sqrt{x^2+3s^2}} \implies y=k_2(x+\sqrt{x^2+3s^2})$$ And $k_2$ can be found out as $k_2=\frac{1}{3s}$, giving me $y=\frac{x+\sqrt{x^2+3s^2}}{3s}$. I'm not sure how to proceed further and solve for $u$?
 A: I think it is easier if we write the ODEs in the parametrisation invariant form of the Lagrange-Charpit equations
$$\frac{dx}{u} = \frac{dy}{y} = \frac{du}{x}$$
You correctly identified the first characteristic curve
$$\frac{dx}{u} = \frac{du}{x} \implies u^{2} - x^{2} = C_{2}$$
Now, we can use componendo dividendo on the first and third fractions to get
\begin{align}
\frac{dx}{u} = \frac{du}{x} &= \frac{du+dx}{u+x} \\
&= \frac{d(u+x)}{u+x} \\
&= \frac{dy}{y} \\
\end{align}
and so solving the last equality and using the functional relationship $C_{2} = f(C_{1})$ yields
\begin{align}
\ln(u+x) &= \ln y + C_{1} \\
\implies \frac{u+x}{y} &= C_{1} \\
\implies u^{2} - x^{2} &= f(C_{1}) \\
&= f \left(\frac{u+x}{y}\right)
\end{align}
which you can check solves the PDE. Now you can apply the data $u(x,1) = 2x$.
A: According to your results we have
$$
\cases{
u^2-x^2-3s^2=0\\
y-\frac{x+\sqrt{x^2+3s^2}}{3s}=0
}
$$
or
$$
\cases{
u^2-x^2-3s^2=0\\
(3sy-x)^2-x^2-3s^2=0
}
$$
So solving the first for $s$ and substituting after choosing the feasible one into the second, we have
$$
u = x\frac{3y^2+1}{3y^2-1}
$$
