PDE characteristic equation I'm having trouble with this IVP PDE problem.
$uu_x +yu_y=x$
with initial conditions $x=s$, $y=s$, and $u=2s$
I get $dx/dt=u$, $dy/dt=y$, and $du/dt=x$
From there I get $x = ut+s$, $y=se^t$, and $u=xt+2s$
I tried to solve for t and s in terms of x and y then express $u$ in these terms but every time I try to solve for t and s the formulas end up incredibly messy and don't work out. Am I setting this up correctly? I tried looking up similar problems but they are always simpler examples that work out or they solve the question in a different method we haven't learned.
 A: $$uu_x+yu_y=x$$
Solving thanks to the method of characteristics :
The set of ODEs for the characteristics equations is 
$\quad \frac{dx}{u}=\frac{dy}{y}=\frac{du}{x}$
From $\quad \frac{dx}{u}\frac{du}{x}\quad\to\quad xdx=udu\quad\to\quad u^2-x^2=c_1\quad$  which is a first characteristic equation.
From $\quad \frac{dx}{u}=\frac{dy}{y}\quad\to\quad \frac{dx}{\sqrt{x^2+c_1}}=\frac{dy}{y}\quad\to\quad \ln|\sqrt{x^2+c_1}+x|=\ln|y|+$constant.
$\frac{\sqrt{x^2+c_1}+x}{y}=c_2 \quad\to\quad \frac{u+x}{y}=c_2\quad$ is a second characteristic equation.
The general solution of the PDE expressed on the form of an implicit equation is :
$$\Phi\left((u^2-x^2)\:,\:\left(\frac{u+x}{y}\right)\right)=0$$
any differentiable function $\Phi$ of two variables.
An equivalent form is :
$$\frac{u+x}{y}=F(u^2-x^2)$$
any differentiable function $F$.
An implicit form of general solution of the PDE is :
$$u=-x+yF(u^2-x^2)$$
The function $F$ has to be determined according to the conditions specified in the wording of the question : $\quad x=y=\frac{u}{2}=s$
$2s=-s+sF((2s)^2-s^2) \quad\to\quad 3=F(3s^2)\quad\to\quad F$ is a constant function $=3$.
According to the specified conditions the particular solution of the PDE is :
$$u=-x+3y$$
Final check :
$u_x=-1\quad u_y=3\quad uu_x+yu_y=-(-x+3y)+3y=x\quad$ OK for the PDE.
$x=y=s \quad\to\quad u=-s+3s=2s\quad$ OK for the specified conditions.
A: $uu_x + uu_y=x$ (this is Lagrange's first order linear pde ) hence the subsidary equation is :
$$\frac{dx}{u}=\frac{dy}{y}=\frac{du}{x
}$$
First $$\frac{dx}{u}=\frac{du}{x} \Rightarrow  x\ dx=u\ du \Rightarrow u^2-x^2=C_1$$
Second $$\frac{dx - du}{u-x}=\frac{dy}{y}\Rightarrow y(u-x)=C_2$$
So the general solution is  :$$y(u-x)=f(u^2-x^2)$$
If $x=y=s $ and u =2s which implies that the particular solution is :
$$y(u-x)=3 \Rightarrow u(x,y)=x+\frac{C}{y}$$
To make sure that this is a right solution check :
$$u_x=1  $$$$u_y=\frac{-C}{y^2}$$
Finally$$uu_x+yu_y=x+\frac{C}{y}+y(\frac{-C}{y^2})=x$$
More precisely : "C=0" that's why i left C in the particular solution it doesn't matter .
