<s>[INST] 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

 [/INST] 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.
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