The following PDE: $(x-y)\;\frac{\partial u}{\partial x} + (y-x-u)\;\frac{\partial u}{\partial y} = u$ with $u(x,0) = 1$
satisfies:
a) $u^2(x - y +u) + (y-x-u) = 0$
b) $u^2(x + y +u) + (y-x-u) = 0$
c) $u^2(x - y +u) + (y+x+u) = 0$
d) $u^2(x - y +u) + (y+x-u) = 0$
My attempt:
I tried to solve the following ODE
$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u}$
First I got $d(x+y +u) = 0$ from where I deduced that $x + y + u = c_1$.
Secondly I substituted $u$ in $y-x-u$ to get $y-x+x+y-c_1 = 2y - c_1$ and the solved
$\frac{dy}{2y - c_1} = \frac{du}{u}$ to get $\frac{\sqrt{y - (c_1/2)}}{u} = c_2$. Hence resubstituting $c_1$, we have $$2c_2^2u^2 = y - x -u.$$
The intial value is not helping me because it gives $c_2^2 = -(x+1)/2$ which is not a constant. Also my answer is nowhere close to the options provided?
In fact neither $c_1$ nor $c_2$ are constants. Better said, choosing values for them defines a particular characteristic curve. The boundary conditions (in this case $u(x,0)=1$) give the initial conditions for each characteristic curve. What you are calculated is the equations that along the curve $(x,0)$ (the $y$ axis) the constants $c_1$ and $c_2$ must satisfy (the second one is obtained by the same procedure as th first):
$c_2^2 = -(x+1)/2$
$x+1=c_1$
For both equations to be true they must to be related this way: $2c_2^2=-c_1$ (substituting $x+1$ in the first by the value from the second)
But the constants have to be related in general, so is,
$\dfrac{y - x -u}{u^2}=2c_2^2=-c_1=-(x+y+u)$ or
$u^2(x + y +u) + (y-x-u) = 0$