Question #23828

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

Expert's answer

xu_x + (x+y)u_y = 1.Method of characteristics:

dx/x = dy/(x+y) = du/1

1) du = dx/x

Integrate:

u=ln(x) + C1 => C1 = u - ln(x)

2) 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

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 or in another form:

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.

dx/x = dy/(x+y) = du/1

1) du = dx/x

Integrate:

u=ln(x) + C1 => C1 = u - ln(x)

2) 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

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 or in another form:

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.

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