Answer to Question #23901 in Differential Equations for Matt Lind
Solve the equation xu_x + (x+y)u_y = 1 with the initial condition u(1,y) = y. Is the solution defined everywhere?
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.