Lagrange's First Order PDE Special Case . How Can I solve this Lagrange's First order PDE ?
$$ xu_{x} +(2y+u)u_{y}=4y+2u$$
I have tried to solve this as follows :
$$\frac{dx}{x} = \frac{dy}{2y+u} = \frac{du}{4y+2u} $$
Then  What came into my mind is this
$$ \frac{dy}{dx} = \frac{2}{x}y+\frac{1}{x}u $$
 $$  \frac{du}{dx} = \frac{4}{x}y+\frac{2}{x}u $$
   this is system of ode with variable coefficient which i don't know how to solve it yet ?
Is there any other idea for solving this lagrange's first pde ?
and If not How Can I solve the previous system ?
Thank you ...
 A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\begin{cases}\dfrac{dy}{dt}=2y+u~......(1)\\\dfrac{du}{dt}=4y+2u~......(2)\end{cases}$
$(1)\times2+(2)$ :
$2\dfrac{dy}{dt}+\dfrac{du}{dt}=8y+4u$
$\dfrac{d(2y+u)}{dt}=4(2y+u)$
$\dfrac{d(2y+u)}{2y+u}=4~dt$
$\int\dfrac{d(2y+u)}{2y+u}=4\int dt$
$\ln(2y+u)=4t+c_1$
$2y+u=C_1e^{4t}$
$(1)\times2-(2)$ :
$2\dfrac{dy}{dt}-\dfrac{du}{dt}=0$
$\dfrac{d(2y-u)}{dt}=0$
$2y-u=C_2$
$\therefore\begin{cases}y=\dfrac{C_1e^{4t}+C_2}{4}\\u=\dfrac{C_1e^{4t}-C_2}{2}\end{cases}$
$y(0)=y_0$ , $u(0)=f(y_0)$ :
$\begin{cases}\dfrac{C_1+C_2}{4}=y_0\\\dfrac{C_1-C_2}{2}=f(y_0)\end{cases}$
$\begin{cases}C_1+C_2=4y_0\\C_1-C_2=2f(y_0)\end{cases}$
$\begin{cases}C_1=2y_0+f(y_0)\\C_2=2y_0-f(y_0)\end{cases}$
$\therefore\begin{cases}y=\dfrac{(2y_0+f(y_0))e^{4t}+2y_0-f(y_0)}{4}\\u=\dfrac{(2y_0+f(y_0))e^{4t}-2y_0+f(y_0)}{2}\end{cases}$
$\begin{cases}4y=2y_0(e^{4t}+1)+f(y_0)(e^{4t}-1)\\2u=2y_0(e^{4t}-1)+f(y_0)(e^{4t}+1)\end{cases}$
$\therefore\begin{cases}y_0=\dfrac{2(e^{4t}+1)y-(e^{4t}-1)u}{8e^{4t}}=\dfrac{x^4(2y-u)+2y+u}{8x^4}\\f(y_0)=\dfrac{(e^{4t}+1)u-2(e^{4t}-1)y}{4e^{4t}}=\dfrac{x^4(u-2y)+2y+u}{4x^4}\end{cases}$
Hence $\dfrac{x^4(u-2y)+2y+u}{4x^4}=f\left(\dfrac{x^4(2y-u)+2y+u}{8x^4}\right)$
