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Two different methods give two different answer

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    I was solving a differential equation and got to the step where I had to integrate. I did it two different methods to get two different answers. I was wondering if anyone knows why one method must not work.

    Method 1

     \int \frac{1}{4y} dy

    let  u = 4y

    therefore:

     \frac{du}{dy} = 4

     \frac{dy}{du} = \frac{1}{4}

     dy = \frac{1}{4} du

    so:

     \int \frac{1}{u} \frac{1}{4} du = \frac{1}{4} \int \frac{1}{u} du

     \frac{1}{4} \ln u = \frac{1} {4} \ln (4y)

    Method 2

     \int \frac{1}{4y} dy

     \frac{1}{4} \int \frac{1}{y}

     \frac{1}{4} \ln y

    As you see both gives different answer but I'm sure both methods are correct.

    Thanks,
    Kieron
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    You missed off the '+c'. Your first answer is equivalent to 1/4*ln(4) + 1/4*ln(y).
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    The key is the "+C", which you forgot to add.

    Notice that if g(x) = f(x) + C then \dfrac{dg}{dx} = \dfrac{df}{dx}, so when you integrate two things then you can end up with two different answers, but they will always only differ by a constant.

    Here, \ln 4y = \ln 4 + ln y, so you get \dfrac{1}{4} \ln 4 + \dfrac{1}{4} \ln y. The \dfrac{1}{4}\ln 4 is a constant, and so (if you add your constants of integration) the two answers are actually the same -- this \dfrac{1}{4}\ln 4 can then be absorbed into the constant of integration in your first method to give \dfrac{1}{4}\ln y + C.
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    You've forgotten your constant of integration.
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    (Original post by nuodai)
    The key is the "+C", which you forgot to add.

    Notice that if g(x) = f(x) + C then \dfrac{dg}{dx} = \dfrac{df}{dx}, so when you integrate two things then you can end up with two different answers, but they will always only differ by a constant.

    Here, \ln 4y = \ln 4 + ln y, so you get \dfrac{1}{4} \ln 4 + \dfrac{1}{4} \ln y. The \dfrac{1}{4}\ln 4 is a constant, and so (if you add your constants of integration) the two answers are actually the same -- this \dfrac{1}{4}\ln 4 can then be absorbed into the constant of integration in your first method to give \dfrac{1}{4}\ln y + C.
    I can't believe something so simple drove me crazy. Most of the times the constant doesn't change the answer so I tend to forget to put it.

    Thanks
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    (Original post by Darkening Light)
    I can't believe something so simple drove me crazy. Most of the times the constant doesn't change the answer so I tend to forget to put it.

    Thanks
    Indeed! In regards to forgetting about the constant, try integrating 1/x dx by parts some time, with u = 1/x, v' = 1.
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    the +c my friend!
    i forgot it when integrating arcsin and i got two different answers by two methods aswell.
    not to worry, just an easy mistake

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Updated: May 10, 2012
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