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Bad method for finding potential function for exact ODE

  1. Nov 2, 2012 #1
    I'm sorry this is going to sound kind of confusing and vague at first but stick with me! I remember a physicsforum thread from long back in which a student posted a test they'd been given back where the instructor marked them off and they argued they were right. The test question was to solve an exact differential equation, and the discussion in the thread seemed to side with the student, claiming that apparently the teacher had found the potential function using a common, albeit not well defined method and it ended up causing a problem in his/her solution. The student on the other hand had solved it using the well defined method and come up with the right solution.


    I have been worried ever since then because this method, which works most of the time, apparently can given problems in some case, and it's the method I was taught in my class. But now I can't remember now which method it is and I can't find the old thread! I'm not looking for a link to an old thread, I just want someone to tell me if my "method" is the well defined one, or the one that will give problems some times.


    When I find an ODE of the form Mdx+Ndy=0 is exact, I know this because My = Nx and there exists a potential function F such that the total diferential of F is Mdx+Ndy.


    I always start out by integrating ∫Mdx so I know F = ∫Mdx + h(y)

    Then I take the partial of F with respect to y, noting that this should equal N itself, so I can figure out h'(y), then integrate and get h(y) and thus F.

    Here's an example:

    Take the ODE: (2xy-3x2)dx + (x2-2y)dy = 0

    Let M=(2xy-3x2), N=(x2-2y)

    My=2x = Nx => this is an exact ODE and there exists potential function F such that its total differential equals Mdx+Ndy


    To find the potential function I start arbitrarily with M

    ∫Mdx = x2y-x3+g(y)

    => F = x2y-x3+g(y)

    => Fy=x2+g'(y)

    But this should equal N so we can determine that g'(y)=-2y => g(y) = -y2 + C


    So now we can plug this all back together and get

    F = x2y-x3 -y2 + C


    Now in this case, I got the right solution. But is this method well defined? I just don't want to be relying on a method that will give me errors on certain problems.
     
  2. jcsd
  3. Nov 3, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    That method is perfectly correct.
     
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