Answer to Question #23828 in Differential Equations for Matthew Lind
dx/x = dy/(x+y) = du/1
1) du = dx/x
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)
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
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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