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Math Help - [SOLVED] find integrating factor and solve the equation 2

  1. #1
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    [SOLVED] find integrating factor and solve the equation 2

    Question:

    (1 - x - z)dx + dz = 0

    let M = (1 - x - z) ; N = 1

    dM/dz = -x ; dN/dx = 0
    hence, not exact.

    integrating factor...

    1/N[ dM/dz - dN/dx ] = 1[-x]
    .................................= -x

    e^ ∫-x dx = e ^ [(-x^2)/2]


    times integrating factor with DE..

    {e ^ [(-x^2)/2] -x e ^ [(-x^2)/2] - z e ^ [(-x^2)/2]}dx +
    e ^ [(-x^2)/2]dz = 0

    dM/dz = -e ^ [(-x^2)/2] ; dN/dx = -xe ^ [(-x^2)/2]

    the problem is, i calculated the integrating factor.. but, the equations still not the exact calculation.
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  2. #2
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    Quote Originally Posted by nameck View Post
    Question:

    (1 - x - z)dx + dz = 0

    let M = (1 - x - z) ; N = 1

    dM/dz = -x ; dN/dx = 0
    hence, not exact.

    integrating factor...

    1/N[ dM/dz - dN/dx ] = 1[-x]
    .................................= -x

    e^ ∫-x dx = e ^ [(-x^2)/2]


    times integrating factor with DE..

    {e ^ [(-x^2)/2] -x e ^ [(-x^2)/2] - z e ^ [(-x^2)/2]}dx +
    e ^ [(-x^2)/2]dz = 0

    dM/dz = -e ^ [(-x^2)/2] ; dN/dx = -xe ^ [(-x^2)/2]

    the problem is, i calculated the integrating factor.. but, the equations still not the exact calculation.
    (1 - x - z)\,dx + dz = 0

    1 - x - z + \frac{dz}{dx} = 0

    \frac{dz}{dx} - z = x - 1.


    Now the integrating factor is e^{-1\,dx} = e^{-x}.


    Multiplying through by the integrating factor gives:

    e^{-x}\,\frac{dz}{dx} - e^{-x}z = e^{-x}(x - 1)

    \frac{d}{dx}(e^{-x}z) = e^{-x}(x - 1)

    e^{-x}z = \int{e^{-x}(x - 1)\,dx}.


    Now using integration by parts with u = x - 1 and dv = e^{-x}...


    e^{-x}z = -e^{-x}(x - 1) - \int{-e^{-x}\,dx}

    e^{-x}z = -e^{-x}(x - 1) - e^{-x} + C

    z = -(x - 1) - 1 + Ce^x

    z = -x + 1 - 1 + Ce^{x}

    z = -x + Ce^{x}

    x + z = Ce^{x}

    (x + z)e^{-x} = C.
    Last edited by Prove It; March 27th 2010 at 07:58 PM.
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  3. #3
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    Prove It....

    i kinda stuck here.. the answer given is (x + z)e^-x = C
    which is not same as the answer u calculated above..
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  4. #4
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    Are you sure it's not (-x + z)e^{-x} = C?
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  5. #5
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    got it!!
    from here...
    [tex]

    e^{-x}z = -e^{-x}(x - 1) - e^{-x} + C


    i got the answer...
    but.. the way u solve this equation is easier than what i've been told in my lecture.. i need to integrate, then differentiate, the compare the value of constant, C.. which is more confusing and the chance for me to make mistakes is high..
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  6. #6
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    Quote Originally Posted by nameck View Post
    got it!!
    from here...
    [tex]

    e^{-x}z = -e^{-x}(x - 1) - e^{-x} + C


    i got the answer...
    but.. the way u solve this equation is easier than what i've been told in my lecture.. i need to integrate, then differentiate, the compare the value of constant, C.. which is more confusing and the chance for me to make mistakes is high..
    Ah I see, I forgot about the negative... I'll edit now...
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