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Simple Differential Equation - Help Me Please

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    Question:

    Given that  \displaystyle y = u^{-1}

    And  \displaystyle \dfrac{du}{dx} - ux^{-1} = -x^{-1}\ln x

    And  \displaystyle y(1)=\frac12

    Find  \displaystyle y=f(x)

    My pdf textbook is notorious for getting things wrong so I would appreciate it if someone could verify my answer, I won't post my full working but:

    Integrating factor:  \displaystyle I = x^{-1}

    Hence  \displaystyle ux^{-1} = \int -x^{-2}\ln x \ dx

    After using integration by parts on the RHS:  \displaystyle ux^{-1} = -x^{-1}\ln x - x^{-1} +C

    Now I get  \displaystyle u = -\ln x -1 + C

    And the book gets  \displaystyle u = \ln x +1 +C

    Without solving for y, which is just the reciprocal, can anyone confimr who is correct ?

    Thnx
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    the book is correct.

    the integration by parts is:

    Integral[ {-1/x^2}{Log[x]} = - (-1/x)(Log[x]) + Integral[(-1/x)(1/x)]dx

    = 1/x+(log[x])/x +C =u/x
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    Sorry, but you are both wrong!

    The correct answer is,

    u = ln(x) - 1 + Cx

    In your IBP, you should have ux^{-1}=+x^{-1}ln(x)-x^{-1}+C

    If you work out \frac{du}{dx}-ux^{-1}=-x^{-1}ln(x)

    using my answer, you will see that it satisfies the DE.
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    (Original post by steve10)
    Sorry, but you are both wrong!

    The correct answer is,

    u = ln(x) - 1 + Cx

    In your IBP, you should have ux^{-1}=+x^{-1}ln(x)-x^{-1}+C

    If you work out \frac{du}{dx}-ux^{-1}=-x^{-1}ln(x)

    using my answer, you will see that it satisfies the DE.
    My mistake was that I took the -1 out of the bracket and forgot to put it back in,
    The correct answer is  u = \ln x +1 +cx .
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    Some really wrong answers on here.. I agree with most of what you've said and Steve10 is the closest.

    The correct answer would be u=ln(x)+x+cx
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    (Original post by member910132)
    My mistake was that I took the -1 out of the bracket and forgot to put it back in,
    The correct answer is  u = \ln x +1 +cx .
    I can confirm this is the correct answer
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    my answer, equated to u/x , then multiplied through by x, is correct.

    (k = 1) and y=1/(x+Log(x)+1)

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Updated: May 17, 2012
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