for the cauchy problem , determine unique solution ,or no solution , or infinitely many solution for the cauchy problem , determine unique solution ,or no solution , or infinitely many solution 
$u_x-6u_y=y$ with the date $u(x,y)=e^x$  on the line $y=-6x+2$
My attempt:
given $u_x-6u_y=y$ then $\frac{dx}{1}=\frac{dy}{-6}=\frac{du}{u}$
there fore the general solution  $\phi(6x+y, y^2+12u)=0$
and hence implicit  equation $y^2+12u=F(6x+y)$
but i cant go for further

 A: given $u_x-6u_y=y$ then $\frac{dx}{1}=\frac{dy}{-6}=\frac{du}{y}\quad$ not $=\frac{du}{u}\quad$(typo).
$y^2+12u=F(6x+y)\quad$ is OK.
Condition $u(x,y)=e^x$  on the line $y=-6x+2\quad\to\quad y^2+12e^x=F(2)=$constant is impossible. Thus, there is no solution.
Condition $u(x,y)=1$ on the line $y=-x^2\quad\to\quad x^4+12=F(6x-x^2)\quad$ allows to determine a function $F(X)= 12+(3\pm\sqrt{9-X})^4$
$$ u(x,y)=\frac{1}{12}\left(-y^2+ 12+\left(3\pm\sqrt{9-(6x-x^2)}\right)^4\right)$$
Condition $u(x,y)=-4x$ on the line $y=-6x\quad\to\quad (-6x)^2+12(-4x)=F(0)=$constant is impossible. Thus, there is no solution.
A: The general rule for hyperbolic PDEs in two variables is that on any boundary you are allowed as many boundary conditions as there are characteristics entering the domain. This question is about the special case where the domain boundary coincides with a characteristic. On such a boundary we may only prescribe boundary data that satisfies the characteristic equation. That is not the case here, so there is a conflict, and therefore no solution.
