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An Exact Equation

  1. Nov 22, 2004 #1
    I've got a problem with exact equations here, the question I've got is,

    [tex] x \frac {dy}{dx} = 2xe^x-y+6x^2 [/tex]

    sp, i put it in the form,

    [tex] x dy-(2xe^x-y+6x^2) dx = 0 [/tex]

    [tex] x dy+(-2xe^x+y-6x^2) dx = 0 [/tex]

    the equation would be exact as,

    [tex]\frac {\partial M}{\partial x}=1[/tex]

    [tex] \frac {\partial N}{\partial y} =1 [/tex]

    But when I integrate M wrt. y and N wrt x I get totally different answers. So which one do I follow? Thanks.

    : )
  2. jcsd
  3. Nov 23, 2004 #2
    Since the equation is exact, there is a function F(x,y) such that
    M = The derivative of F with respect to y ... (1)

    N= The derivative of F with respect to x. ...(2)
    So, F = The integral of M with respect to y ( keeping x constant) + a function of x (which I'll call g(x)).
    Differentiating with respect to x,
    N = d/dx ( Int.Mdy) + g'(x). Now g can be ( in principle) found out.

    Are you sure you included g in the integration?
    I'm,with great respect,
  4. Nov 23, 2004 #3
    thanks for the help, i forgot to include the g(x) as you have said. I don't think I'll start another thread but I've now currently got a problem. It has something to do with the linear equation.

    [tex] x^-^4 \frac {dy}{dx} 4x^-^5 y = xe^x [/tex]

    I don't know how to simplify this to,

    [tex] \frac {d}{dx} (x^-^4 y) = xe^x [/tex]


    (I'm now at my friend's house and the book is not with me, so I'll dig out the original question soon.)
  5. Nov 24, 2004 #4


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    Erm, I'm not convinced you wrote this out correctly (I'm guessing there should be - between the dy/dx and the 4x^-5 but if you did:

    [tex] x^{-4} \frac {dy}{dx} 4x^{-5} y = xe^x [/tex]

    [tex]\frac{dy}{dx} \frac{1}{x^9} y = xe^x[/tex]

    [tex]y\frac{dy}{dx} = x^{10} e^x[/tex]

    [tex]\int y \frac{dy}{dx} dx = \int x^{10} e^x[/tex]

    Then all you need to do is use by-parts a lot. But assuming you meant there to be a minus there:

    [tex]x^{-4} \frac {dy}{dx} - 4x^{-5} y = xe^x[/tex]

    Notice the LHS takes the form d/dx (uv) = uv'+u'v, where u=x^-4 and v=y. So rewriting:

    [tex]\frac{d}{dx} \left( x^{-4} y \right) = xe^x[/tex]
  6. Nov 24, 2004 #5
    I'm sorry, there was a mistake, it should be,

    [tex] x^-^4 \frac {dy}{dx} - 4x^-^5y =xe^x [/tex]

    there should be a minus sign.....

    I've finally understood, thanks alot....

    : )
    Last edited: Nov 24, 2004
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