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Is M(x,y)dx + M(y,x)dy Always Exact?

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data

    I thought that this was an interesting question.

    (a) Show that [itex](x^3 + y)dx + (y^3 +x)dy = 0\qquad(1)[/itex] is exact.
    (b)More generally, is [itex]M(x,y)dx + M(y,x)dy\qquad(2)[/itex] exact? Explain.

    2. Relevant equations
    Test for exactness: [tex]\left(\frac{\partial{M}}{\partial{y}}\right)_x=\left(\frac{\partial{N}}{\partial{x}}\right)_y[/itex]

    3. The attempt at a solution
    (a) Applying the test is simple enough. 1 = 1. Exact.

    (b) Now this is how I am thinking about (b). Please correct me if I am wrong. M is simply a rule. It is in the form M(x1,x2). It tells us how to operate on whatever is in the x1 and x2 spot. In equation (2), in the first term, x1 = x and x2 = y , and in the second term x1 = y and x2 = x. Wouldn't this imply that the equation will always be exact? Is there a way to show that it is or isn't?

    Any thoughts?
  2. jcsd
  3. Sep 25, 2010 #2


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    Suppose there is a function z=f(x,y). Your M and N are partial derivatives of z w.r.t x and y

    [tex]M = \left(\frac{\partial{z}}{\partial{x}}\right), N = \left(\frac{\partial{z}}{\partial{y}}\right)[/tex]

    By taking a derivative of M with respect to y, and taking derivative of N with respect to x, you are essentially testing for exactness. It is an exact differential equation if and only if:

    [tex]\left(\frac{\partial{}}{\partial{y}}\right) \left(\frac{\partial{z}}{\partial{x}}\right) = \left(\frac{\partial{}}{\partial{x}}\right) \left(\frac{\partial{z}}{\partial{y}}\right)[/tex]

    Your objective is to find z.

    I think to make it clear you need an example:

    Suppose there are 2 functions, z1= 2*x^2 + y^2, z2 = x^2 + 2*y^2
    M(z1) = 4x, N(z1) =2y M(z2) = 2x, N(z2) = 4y
    My = 0, Nx = 0 My(z2) = 0, Nx(z2) = 0

    So if I give you (4x)dx + (2y)dy = 0, your solution will be integral(M(z1))dx + integral(N(z1))dy =2x^2 + y^2 , and

    if I give you (2x)dx + (4y)dy = 0, your solution will be integral(M(z2))dx + integral(N(z2))dy = x^2 + 2y^2.

    This was only possible because both M and N were describing the same function z
    Last edited: Sep 25, 2010
  4. Sep 25, 2010 #3
    Hi cronxeh :smile: I understand the test for exactness and how to find F(x,y) if it is exact. The question in part (b), is if equations of the form (2) are always exact. I am not sure how to apply what you have posted to part (b).
  5. Sep 25, 2010 #4


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    So wait you saying part b is not a typo? You actually want to find whether M(x,y)dx + M(y,x)dy = c is always exact?
  6. Sep 25, 2010 #5
    You can try to see what happens by developing M(x,y) into Taylor series around (0,0) and try on [tex]M(x,y)=x^my^n[/tex]. Once you are done, use [tex](x_0,y_0)[/tex] instead of [tex](0,0)[/tex]. This may give you some insight and intuition to start with.
    Last edited: Sep 25, 2010
  7. Sep 25, 2010 #6
    Indeed it is not a typo.
  8. Sep 25, 2010 #7
    Notice that when you write

    [tex]\frac{\partial M(x,y)}{\partial x}[/tex] it can be written as well as

    [tex]\left(\frac{\partial M(s,t)}{\partial s}\right)_{s=x,t=y}[/tex]


    [tex]\left(\frac{\partial M(t,s)}{\partial t}\right)_{t=x,s=y}[/tex]

    It's all the same. Now, you can do it.
  9. Sep 25, 2010 #8

    D H

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    Hint (Big hint): Find a counterexample. It's not hard.
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