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Math Help - Exact Differential Equation

  1. #1
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    Exact Differential Equation

    I am told to verify that the given differential equation is exact; then solve it:
    (x + tan-1y)dx + ((x + y)/(1 + y^2))dy=0

    So far I have found the exactness and have gotten down to
    F= (x^2) + 2xtan^(-1) (y) + g(y)

    I am then trying to find g(y) and to do this do I have to set
    F= (x^2) + 2xtan^(-1) (y) + g(y) = ((x + y)/(1 + y^2))
    If I do, how in the world do I cancel things out to get the anwer!

    The answer is suppose to be
    (x^2) + 2xtan^(-1) (y) + ln(1+y^2)

    Please help

    Thanks
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    I am told to verify that the given differential equation is exact; then solve it:
    (x + tan-1y)dx + ((x + y)/(1 + y^2))dy=0

    So far I have found the exactness and have gotten down to
    F= (x^2) + 2xtan^(-1) (y) + g(y)

    I am then trying to find g(y) and to do this do I have to set
    F= (x^2) + 2xtan^(-1) (y) + g(y) = ((x + y)/(1 + y^2))
    If I do, how in the world do I cancel things out to get the anwer!

    The answer is suppose to be
    (x^2) + 2xtan^(-1) (y) + ln(1+y^2)

    Please help

    Thanks
    If M dx + N dy = 0 is exact then M_y = N_x (which you did). If so, then F exist such that

    F_x = m,\;\;\;F_y = N

    so in your case

    F_x = x + \tan^{-1}y,\;\;\;F_y = \frac{x+y}{1+y^2}

    From the first we obtain

    F = \frac{x^2}{2} + x\tan^{-1}y + g(y)

    Now substitute this into the above expression for Fy, i.e.

    F_y = \frac{x+y}{1+y^2}

    This will get you an equation for g(y) or g' in particular.
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