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Exact Differential Equations

  1. Jul 20, 2014 #1
    I thought I got this problem wrong, but I think I have it right now. It turned out that when I was taking the derivative of e^xy2with respect to y, I forgot that you're supposed to multiply by 2xy (the derivative of xy2), not just x.

    1. The problem statement, all variables and given/known data

    (y2* e^xy2 + 4x3 dx + (2xy * e^xy2 - 3y2 dy = 0

    2. Relevant equations

    3. The attempt at a solution

    First, I test to make sure it's exact. When I differentiate the first with respect to y, and the second with respect to x, I get 2xy3 * e^xy2 + 2y*e^xy2 for both, so it is indeed exact.

    When I integrate (y2* e^xy2 + 4x3 dx, I get e^xy2 + x^4 + f(y)

    The derivative of e^xy2 + x^4 + f(y) with respect to y is 2xy*e^xy2 + f'(y), and f'(y) is (2xy * e^xy2 - 3y2 dy), so I integrate 3y2 with respect to y, and get y3

    So my final function is f(x,y) = e^xy2 + x4 + y3
     
    Last edited: Jul 20, 2014
  2. jcsd
  3. Jul 20, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    I think you meant the equation (y2* e^(xy2) + 4x3)dx + (2xy * e^(xy2) - 3y2 )dy = 0. Do not forget parentheses!

    Excellent! But you need to include the integration constant, so f(x,y) = e^(xy2) + x4 + y3 + C

    ehild
     
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