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Math Help - Inexact Differential Equation

  1. #1
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    Exclamation Inexact Differential Equation

    I need help with this:

    Given: [x - (sec y)(ln x)] dy + [tan y – (y/x)sec y] dx = 0
    Required: General Solution
    ----------------------------------------------------------------------

    Suppose that M = tan y – (y/x) sec y and N = x – (sec y) (ln x)
    Then, partial derivative of N wrt x = 1 – ((sec y)/x)
    And partial derivative of M wrt y = sec y (sec y – ((y tan y)/x) - 1/x)
    f(y) = - 1/M (sec y (sec y – ((y tan y)/x) – 1/x))
    from there, to get the integrating factor,
    f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
    and then, i’m kind of stuck.
    Last edited by vividlyxinsane; July 5th 2009 at 11:28 PM.
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  2. #2
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    Quote Originally Posted by vividlyxinsane View Post
    I need help with this:

    Given: [x - (sec y)(ln x)] dy – [tan y – (y/x)sec y] dx = 0
    Required: General Solution
    ----------------------------------------------------------------------

    I can't seem to come up with the right answer.

    Help? =]
    are you sure that - is not + ?
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    Quote Originally Posted by NonCommAlg View Post
    are you sure that - is not + ?

    oh yeah, it's positive. sorry.

    back to the problem,

    Suppose that M = tan y –(y/x) sec y and N = x – (sec y) (ln x)
    Then, partial derivative of N wrt x = 1 – ((sec y)/x)
    And partial derivative of M wrt y = sec y (sec y – ((y tan y)/x) - 1/x)
    f(y) = - 1/M (sec y (sec y – ((y tan y)/x) – 1/x))
    from there, to get the integrating factor,
    f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
    and then, i’m kind of stuck.
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  4. #4
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    Quote Originally Posted by vividlyxinsane View Post
    oh yeah, it's positive. sorry.

    back to the problem,

    Suppose that M = tan y –(y/x) sec y and N = x – (sec y) (ln x)
    Then, partial derivative of N wrt x = 1 – ((sec y)/x)
    And partial derivative of M wrt y = sec y (sec y – ((y tan y)/x) - 1/x)
    f(y) = - 1/M (sec y (sec y – ((y tan y)/x) – 1/x))
    from there, to get the integrating factor,
    f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
    and then, i’m kind of stuck.
    just multiply both sides of your original differential equation by \cos y and you'll have an exact equation.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    just multiply both sides of your original differential equation by \cos y and you'll have an exact equation.

    cos y is the integrating factor?

    before i get there, f(y) has to be purely in terms of y. i'm stuck at:

    f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
    i'm kind of clueless on what to do next.
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  6. #6
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    Quote Originally Posted by vividlyxinsane View Post
    cos y is the integrating factor?

    before i get there, f(y) has to be purely in terms of y. i'm stuck at:

    f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
    i'm kind of clueless on what to do next.
    since \sec^2y - 1 = \tan^2y, your function becomes: \frac{x\tan^2 y - y \sec y \tan y}{x\tan y - y \sec y}=\tan y=f(y).
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  7. #7
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    Quote Originally Posted by NonCommAlg View Post
    since \sec^2y - 1 = \tan^2y, your function becomes: \frac{x\tan^2 y - y \sec y \tan y}{x\tan y - y \sec y}=\tan y=f(y).
    so that was what i was missing. but shouldnt the IF = sec y?

    since the integral of tan y is -ln(cos y) or ln(sec y)?
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  8. #8
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    Quote Originally Posted by vividlyxinsane View Post
    so that was what i was missing. but shouldnt the IF = sec y?

    since the integral of tan y is -ln(cos y) or ln(sec y)?
    the function that you called f(y) is equal to \frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{M} not \frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}. if \mu is an integrating factor which is purely a function of y, then: \frac{d \mu}{\mu}= \frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}dy= -f(y)dy=-\tan y \ dy. so \mu=\cos y.
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    the function that you called f(y) is equal to \frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{M} not \frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}. if \mu is an integrating factor which is purely a function of y, then: \frac{d \mu}{\mu}= \frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}dy= -f(y)dy=-\tan y \ dy. so \mu=\cos y.
    oh, right.. f(y) IS NEGATIVE.
    i got it.
    THANKS A LOT! =]
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