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

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

    y dx + x ln x dy = 0 ; x > 0


    my integrating factor is x..


    so.. multiply with DE,


    xy dx + x^2 ln x dy = 0


    let M = xy ; N = x^2 ln x


    dM/dy = x ; dN/dx = x + 2x ln x


    the problem is.. i didn't get the exact equation after multiply the integrating factor.. i've double checked my integrating factor.. but.. still can't get the exact equation...
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  2. #2
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    Quote Originally Posted by nameck View Post
    y dx + x ln x dy = 0 ; x > 0


    my integrating factor is x..


    so.. multiply with DE,


    xy dx + x^2 ln x dy = 0


    let M = xy ; N = x^2 ln x


    dM/dy = x ; dN/dx = x + 2x ln x


    the problem is.. i didn't get the exact equation after multiply the integrating factor.. i've double checked my integrating factor.. but.. still can't get the exact equation...
    Dear nameck,

    Take the integrating factor as \frac{1}{x} and try.

    Hope this will help you.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear nameck,

    Take the integrating factor as \frac{1}{x} and try.

    Hope this will help you.
    yes.. it works!!
    but.. why is 1/x and not x?
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    Quote Originally Posted by nameck View Post
    yes.. it works!!
    but.. why is 1/x and not x?
    Dear nameck,

    It just occured me that this integrating factor would work. If somebody could tell us a mathamatical way of getting the integrating factor for any first order differential equation that would be helpful is'nt??
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  5. #5
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    If I am given a DE like this, I usually convert it to a first order linear if possible.


    y\,dx + x\ln{x}\,dy = 0

    y + x\ln{x}\,\frac{dy}{dx} = 0

    \frac{dy}{dx} + \frac{1}{x\ln{x}}\,y = 0.


    Now this is first order linear. So the integrating factor is

    e^{\int{\frac{1}{x\ln{x}}\,dx}} = e^{\int{\frac{1}{u}\,du}} after making the substitution u = \ln{x}

     = e^{\ln{u}}

     = e^{\ln{(\ln{x})}}

     = \ln{x}.


    So multiplying through by the integrating factor gives:

    \ln{x}\,\frac{dy}{dx} + \frac{1}{x}\,y = 0

    \frac{d}{dx}(y\ln{x}) = 0

    y\ln{x} = \int{0\,dx}

    y\ln{x} = C

    y = \frac{C}{\ln{x}}.
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