Answer to Question #23901 in Differential Equations for Matt Lind

Question

Solve the equation xu_x + (x+y)u_y = 1 with the initial condition u(1,y) = y. Is the solution defined everywhere?

Answer 1

Method of characteristics:
dx/x = dy/(x+y) = du/1
From 1st and 3rd:
du = dx/x
Integrate:
u=ln(x) + C1 => C1 = u - ln(x)
From 1st and 2nd:
dx/x = dy/(x+y)
dx(x+y) = dy*x
dx*x = dy*x - dx*y
Divide to x^2:
dx/x = (dy*x - dx*y)/x^2 - total derivative of (y/x)
dx/x = d(y/x)
Integrate:
ln(x) = y/x +C2
C2 = ln(x) - y/x
Therefore, the complete integral of the equation:
F ( u - ln(x), ln(x) - y/x) = 0
expressing u:
u = ln(x) + f ( ln(x) - y/x) where f - some differentiable function
The initial condition u(1,y) = y:
u (1, y) = ln(1) + f ( ln(1) - y/1) = f(y) = y
Therefore:
u (x, y) = ln(x) + ln(x) - y/x = y/x
The solution is undefined for x=0.
Answer: u (x, y) = y/x, the solution is undefined for x=0.