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Finding a function given its partial derivatives, stuck on finding g'(x)

  1. Aug 7, 2012 #1
    Hi all,

    I have the following partial derivatives

    ∂f/∂x = cos(x)sin(x)-xy2

    ∂f/∂y = y - yx2

    I need to find the original function, f(x,y).

    I know that df = (∂f/∂x)dx + (∂f/∂y)dy

    and hence

    f(x,y) = ∫∂f/∂x dx + g(y) = -1/2(x2y2+cos2(x)) + g(y)

    Then to find g(y) I took the partial derivative of f(x,y) that I just found wrt y and equate to the original given ∂f/∂y

    i.e -x2y +g'(y) = y-x2
    which is a first order differential equation of the form

    dg/dy = - (y+x2)/(x2y)

    and now I am not sure how to proceed and solve this DE

    Mitch
     
  2. jcsd
  3. Aug 7, 2012 #2

    micromass

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    That's a typo, no? It should be

    [tex]-x^2y+g^\prime(y)=y-yx^2[/tex]
     
  4. Aug 7, 2012 #3
    oh you're right, I must have typed it out incorrectly from my notes.

    so it becomes

    dg/dy = y

    ∴ ∫dg = ∫y dy

    so g = y2/2

    does that sound about right?
     
  5. Aug 7, 2012 #4

    haruspex

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    Well, it satisfies the original equations, right? But don't forget to allow a constant term.
     
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