# $(x - y) u_x + (y - x - u) u_y = u$ and $u(x , 0) = 1$

The Solution of the initial value problem

$(x - y) u_x + (y - x - u) u_y = u$ and $u(x , 0) = 1$ , satisfies

1) $u^2(x + y+ u) +(y - x- u) = 0$

2)$u^2(x- y + u) +(y - x - u) = 0$

3)$u^2(x- y + u) - (y + x + u) = 0$

4) $u^2(y - x + u) +(y + x - u) = 0$

I have applied Lagranage's method. I got $u + x + y = c_1$ and $c_2 = -1$

We have to ways, check the proposed solutions in the equation or solve it. You try to solve it. The second condition for the characteristics is not correct.

The auxiliar system of equations is:

$\dfrac{dx}{x-y}=\dfrac{dy}{y-x-u}=\dfrac{du}{u}$

Substracting the first ratio to minus the third:

$\dfrac{-dx-du}{y-x-u}=\dfrac{dy}{y-x-u}$

$dx+dy+du=0\implies x+y+u=c_1$

with the second and third and using the last equation $-x-u=y-c_1$

$\dfrac{dy}{2y-c_1}=\dfrac{du}{u}$

$(1/2)\ln\vert 2y-c_1\vert=\ln\vert u\vert+k$

$c_2(y-x-u)=u^2$

$f(x+y+u)(y-x-u)=u^2$; for the general solution ($f$ some single argument, regular function)(right now we can see the only plausible solution among the proposed ones is 1). Now, the boundary conditions, $u(x,0)=1$, impose:

$f(x+1)(-x-1)=1\implies f(x+1)=-1/(x+1)\implies f(x)=-1/x$

Finally $-\dfrac{y-x-u}{x+y+u}=u^2$

The solution is 1) $u^2(x+y+u)+(y-x-u)=0$