cauchy problem pde $$(6u+2y)U_x +(3x-6u)U_y+3x+2y=0 ,  x>0 , y>0  $$   $$U(x,0)=x$$
question: Find the solution for the initial value problem.
$\Gamma=(r,0,r)$  I parameterized with r
$$\frac{dx(r,s)}{ds}=6u+2y\\\frac{dy(r,s)}{ds}=3x-6u\\\frac{du(r,s)}{ds}=-3x-2y$$ 
However, i am stuck on the question as im finding it difficult to solve the Characteristic equations.
I would greatly appreciate if someone could push me into the right direction by following the method i am using for consistency.
Thank you
 A: The solution is (calculus below) :
$$u(x,y)=\frac{3}{5}(x+y)+\frac{2}{15}\sqrt{9x^2+108xy+84y^2}$$

The only small difficulty is to solve the ODE : $y'=\frac{-6c_1+9x+6y}{6c_1-6x-4y}$
The usual method consists in a change of variables: $x=t+A$ and $y=s+B$. Determine the constants $A$ and $B$ so that the ODE becomes homogeneous. Then, change of variable $z=\frac{s}{t}$. That way, you get to a separable ODE.
A: You found that you have to solve the first order linear system
$$
\frac{d}{ds}\begin{bmatrix}x\\y\\z\end{bmatrix}
=
\begin{bmatrix}0&2&6\\3&0&-6\\-3&-2&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
for the characteristic curves. Usually this is done via finding eigenvalues and eigenvectors.
You can reduce the dimension by observing that summng up the equations $\frac{d}{ds}(x+y+z)=0$, that is $x+y+z=a=const.$ or $z=a-x-y$.
The reduced system is
$$
\frac{d}{ds}\begin{bmatrix}x\\y\end{bmatrix}
=
\begin{bmatrix}-6&-4\\9&6\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix}
+
\begin{bmatrix}6a\\-6a\end{bmatrix}.
$$
This can be further reduced as the system matrix has rank one. Finding a left null vector one gets 
$$
\frac{d}{ds}(3x+2y)=(6y+18z)+(6y-12z)=6(x+y+z)=6a
$$
so that $3x+2y=6as+b$, etc.
