$$(6u+2y)U_x +(3x-6u)U_y+3x+2y=0 , x>0 , y>0 $$ $$U(x,0)=x$$
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
On
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.
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.