Therefore, the complete integral of the equation: F ( 1/u + x, x - y) = 0 or in another form: u = ( f (x- y) - x )^-1 where f - some differentiable function. The initial condition u(x,0) = 1. u(x, 0) = ( f (x) - x )^-1 = 1 => f(x) = x+1 u(x,y) = 1/(x-y+1-x) = 1/(-y+1) = 1/(1-y) Answer: u (x, y) = 1/(1-y)
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