Need help with a quasi-linear PDE Let be the following quasi-linear PDE:
$(xu+y)\frac{\partial u}{\partial x} + (x+uy)\frac{\partial u}{\partial y} = 1 - u^2$. 
I wrote the characterstic system : $\frac{dx}{xu+y} = \frac{dy}{x+uy} = \frac{du}{1-u^2}$. And from here, how can I go on ?
 A: It's OK for the chatacteristics system. Afterwards, it's a matter of combination of the differentials to put them on the form of exact differentials (below) :
 
$F(X,Y)$ is any differentiable function of two variables, with $X=(x+y)(u-1)$ and $Y=(x-y)(u+1)$.
A: The $u$ equation give $$ \frac{du}{1-u^2} = dt \,\,\,\, \implies \,\,\,\, \frac 1 2 \left( \frac{1}{1+u} - \frac 1{1-u} \right)du = dt$$ which results in $$\log \left( \frac{1+u}{1-u} \right) = 2t + c$$ or after rearranging: $$u = \frac{e^{2t} - c}{e^{2t} + c}$$ for some $c > 0$. It actually matters what $c$ is specifically; it will be determined by the boundary data. Next $$\frac{dx}{dt} = xu+y, \,\,\,\,\,\,\, \frac{dy}{dt} = yu+x.$$ Put $X = x+y$ and $Y = x-y$. Then $$\frac{dX}{dt} = (u+1)X, \,\,\,\,\,\, \frac{dY}{dt} = (u-1)Y.$$ Using these we can solve for $X,Y$, then $x,y$ (this part will likely get ugly). This will give us $(x,y)$ in terms of $(t,c)$. Solving for $(t,c)$ gives $t(x,y), c(x,y)$ and the final solution will be given by $$u(x,y) = \frac{e^{2t(x,y)} - c(x,y)}{e^{2t(x,y)} + c(x,y)}.$$
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{du}{dt}=1-u^2$ , letting $u(0)=0$ , we have $u=\tanh t$
$\therefore\begin{cases}\dfrac{dx}{dt}=x\tanh t+y~......(1)\\\dfrac{dy}{dt}=x+y\tanh t~......(2)\end{cases}$
$(1)+(2)$ :
$\dfrac{dx}{dt}+\dfrac{dy}{dt}=x+y+(x+y)\tanh t$
$\dfrac{d(x+y)}{dt}=(x+y)(1+\tanh t)$
$\dfrac{d(x+y)}{x+y}=(1+\tanh t)~dt$
$\int\dfrac{d(x+y)}{x+y}=\int(1+\tanh t)~dt$
$\ln(x+y)=t+\ln\cosh t+c_1$
$x+y=C_1e^t\cosh t$
$(1)-(2)$ :
$\dfrac{dx}{dt}-\dfrac{dy}{dt}=y-x-(y-x)\tanh t$
$\dfrac{d(x-y)}{dt}=(x-y)(\tanh t-1)$
$\dfrac{d(x-y)}{x-y}=(\tanh t-1)~dt$
$\int\dfrac{d(x-y)}{x-y}=\int(\tanh t-1)~dt$
$\ln(x-y)=\ln\cosh t-t+c_2$
$x-y=C_2e^{-t}\cosh t$
$\therefore\begin{cases}x=\dfrac{C_1e^t\cosh t+C_2e^{-t}\cosh t}{2}\\y=\dfrac{C_1e^t\cosh t-C_2e^{-t}\cosh t}{2}\end{cases}$
$x(0)=x_0$ , $y(0)=f(x_0)$ :
$\begin{cases}\dfrac{C_1+C_2}{2}=x_0\\\dfrac{C_1-C_2}{2}=f(x_0)\end{cases}$
$\begin{cases}C_1=x_0+f(x_0)\\C_2=x_0-f(x_0)\end{cases}$
$\therefore\begin{cases}x=\dfrac{(x_0+f(x_0))e^t\cosh t+(x_0-f(x_0))e^{-t}\cosh t}{2}\\y=\dfrac{(x_0+f(x_0))e^t\cosh t-(x_0-f(x_0))e^{-t}\cosh t}{2}\end{cases}$
$\begin{cases}x=x_0\cosh^2t+f(x_0)\sinh t\cosh t\\y=x_0\sinh t\cosh t+f(x_0)\cosh^2t\end{cases}$
$\therefore\begin{cases}x_0=x-y\tanh t=x-yu\\f(x_0)=y-x\tanh t=y-xu\end{cases}$
Hence $y-xu=f(x-yu)$
