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Homework Help: Find the function M?

  1. Oct 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the function M such that M(x, y)dx+(x^2-y^2)dy=0 is exact.

    2. Relevant equations
    This is an exact equation.

    3. The attempt at a solution
    Here's my work:

    integral of (x^2-y^2)dy=yx^2-y^3/3+h(x)=C
    2xy+h'(x)=M(x, y)
    But the answer in the book is M(x, y)=2xy+f(x). How do I get to the answer from there?
  2. jcsd
  3. Oct 13, 2014 #2
    By putting h'(x)=f(x)
  4. Oct 13, 2014 #3
    So you have ##M(x,y)=2xy+h'(x)## and the book has ##M(x,y)=2xy+f(x)##. Seems to me that your answer is basically the same as the book's; you just have different names for the the "extra" function of ##x##.
  5. Oct 13, 2014 #4


    Staff: Mentor

    They want a function M such that My = Nx.
    $$\frac{\partial M}{\partial y} = \frac{\partial (x^2 - y^2)}{\partial x} = 2x$$
    If you integrate 2x with respect to y, you get 2xy + some arbitrary function of x. Taking the partial with respect to y of a function that involves only x is similar to taking the derivative of a constant.
  6. Oct 13, 2014 #5


    User Avatar
    Science Advisor

    Let f(x,y) be any function of two variables and assume that x and y are themselves functions of some variable, t. Then we could write f as a function of the single variable t and, by the chain rule, [itex]df/dt= (\partial f/\partial x) dx/dt+ (\partial f/\partial y)dy/dt[/itex]. In terms of "differentials" that is [itex]df= (\partial f/\partial x)dx+ (\partial f/\partial y)dy[/itex].

    An expression of the form [itex]M(x,y)dx+ N(x,y)dy[/itex] is an "exact differential" if and only if there exist a function f such that [itex]\partial f/\partial x= M(x, y)[/itex] and [itex]\partial f/\partial y= N(x, y)[/itex]. Of course, if that is true, we have [itex]\partial M/\partial y= \partial^2 f/\partial x\partial y[/itex] and [itex]\partial N/\partial x= \partial^2 f/\partial y\partial x[/itex]. But those "mixed second partial derivatives" must be equal so to be an "exact differential" we must have [itex]\partial M/\partial y= \partial N/\partial x[/itex]

    Here, [itex]N= x^2- y^2[/itex] so that [itex]\partial N/\partial x= 2x[/itex]. We must have [itex]\partial M/\partial y= 2x[/itex] so that, integrating with respect to y while holding x constant, we have [itex]M= 2xy+ f(x)[/itex]. "f(x)" is the constant of integration- since we integrate with respect to y while holding x constant, that "constant of integration" can be any function of x.
  7. Oct 14, 2014 #6
    Thank you so much, @Mark44!
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