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Differential Equations curves

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data

    Given u(x.y), find the exact differential equation du = 0. What sort of curves are the solution curves u(x,y) = constant? (These are called the level curves of u).

    u = cos(x2 - y2)


    3. The attempt at a solution
    partial derivative du/dx = (-2x)sin(x2 - y2)dx
    partial derivative du/dy = (2y)sin (x2 - y2)dy

    P = (-2x)sin(x2 - y2)dx ; Q = (2y)sin (x2 - y2)dy

    I found that they're not exact differential equations

    so, I'm using integration factors.

    1/Q(P dx - Q dy) = R(x)

    (1/(2y)sin (x2 - y2))((-2x)sin(x2 - y2) - (2y)sin (x2 - y2)) = R(x)

    I get, R(x) = -x/y

    Then,

    F = exp^(R(x))dx

    I get, F = exp^((-x^2)/(2y))

    Then,

    M = FP and N = FQ

    so,

    M = (exp^((-x^2)/(2y)))((-2x)sin(x2 - y2))

    and

    N = (exp^((-x^2)/(2y)))((2y)sin (x2 - y2))

    I then took the partial derivative of M in respect to y, and this is where I am getting stuck.

    I also have to take the partial derivative of N in respect to x.

    Any help would be much appreciated.
     
  2. jcsd
  3. Oct 4, 2015 #2
    At constant u,
    $$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=0$$
    What does this tell you about dy/dx?

    Chet
     
  4. Oct 4, 2015 #3
    I believe it tells me that the partial derivatives are continuous
     
  5. Oct 4, 2015 #4
    it may also tell me that the partials in respect to x and y are equal to u and 0....?
     
  6. Oct 5, 2015 #5
    Checking for exactness,
    P = (-2x)sin(x2 - y2)dx, Q = (2y)sin (x2 - y2)dy

    dP/dy = (4xy)cos(x^2- y^2)
    dQ/dx = (4xy)cos(x^2 - y^2)

    dP/dy = dQ/dx, they are exact!
    So, don't use the "integration factors", just find the general solution.
     
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