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Math Help - Tricky integral

  1. #1
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    Tricky integral

    Whilst working through a question I got to the point where I need to integrate this:

    1/y + (1/k)/(1-y/k) where k is a constant.

    I got ln|y| + 1/k (ln|1-y/k|)

    I know this isn't correct. Could anyone explain why and give me the correct answer?

    Thankyou
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chr91 View Post
    Whilst working through a question I got to the point where I need to integrate this:

    1/y + (1/k)/(1-y/k) where k is a constant.

    I got ln|y| + 1/k (ln|1-y/k|)

    I know this isn't correct. Could anyone explain why and give me the correct answer?

    Thankyou
    to integrate (1/k)/(1 - y/k), do a substitution: u = 1 - y/k

    now continue with that and see where you get. and try to notice a pattern so that you don't have to go through the u-sub all the time. and remember that in general, to integrate 1/something, you don't automatically take the ln|something|, it very much depends on the kind of function "something" is. follow the rules exactly. you have \displaystyle \int \frac 1x~dx = \ln |x| + C. If it doesn't (or can't) look like 1/x (that is one over a variable to the first degree), then you won't be able to integrate it and get ln. what you have doesn't look like 1/x, so you had to change it to look like that. And that would change other things as well.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    \displaystyle \int \dfrac{1}{y} + \dfrac{\frac{1}{k}}{1-\frac{y}{k}}\ dy = \int \dfrac{1}{y} + \dfrac{1}{k-y} \ dy

    I multiplied the second fraction by k/k to simplify.

    Then, I get:

    \displaystyle \int \dfrac{1}{y} + \dfrac{1}{k-y} \ dy = ln|y| - ln|k-y| + c
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  4. #4
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    Quote Originally Posted by chr91 View Post
    Whilst working through a question I got to the point where I need to integrate this:

    1/y + (1/k)/(1-y/k) where k is a constant.

    I got ln|y| + 1/k (ln|1-y/k|)

    I know this isn't correct. Could anyone explain why and give me the correct answer?

    Thankyou
    \int{\frac{1}{y}}dy+\frac{1}{k}\int{\frac{1}{1-\frac{y}{k}}dy}=\int{\frac{1}{y}}dy+\frac{k}{k}\in  t{\frac{1}{k-y}}dy

    u=k-y
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  5. #5
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    Quote Originally Posted by Unknown008 View Post
    \displaystyle \int \dfrac{1}{y} + \dfrac{\frac{1}{k}}{1-\frac{y}{k}}\ dy = \int \dfrac{1}{y} + \dfrac{1}{k-y} \ dy

    I multiplied the second fraction by k/k to simplify.

    Then, I get:

    \displaystyle \int \dfrac{1}{y} + \dfrac{1}{k-y} \ dy = ln|y| - ln|k-y| + c
    Got it.

    Thanks!
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  6. #6
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    Quote Originally Posted by chr91 View Post
    Sorry for being dumb here but why is it -ln|k-y| not + ln|k-y| ?
    Let u = k-y, as suggested above, to see why.
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  7. #7
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    I now need to get this into a function where y = ....

    ln|y| - ln|k-y| = rt + C

    I need to get the function in terms of y I guess?

    ln |y/(k-y)| = rt + C

    y/(k-y) = Ae^rt

    I can't get y on it's own though. There's too many constants!
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  8. #8
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    Quote Originally Posted by chr91 View Post
    I now need to get this into a function where y = ....

    ln|y| - ln|k-y| = rt + C

    I need to get the function in terms of y I guess?

    ln |y/(k-y)| = rt + C

    y/(k-y) = Ae^rt

    I can't get y on it's own though. There's too many constants!
    I don't know where this rt+C is coming from, but...

    \mathrm{ln}|y|-\mathrm{ln}|k-y|=rt+C
    \mathrm{ln}|\frac{y}{k-y}|=rt+C
    \frac{y}{k-y}=Ae^{rt} (if you want it like this, A=e^C)
    y=(k-y)Ae^{rt}
    y=Ake^{rt}-yAe^{rt}
    y+yAe^{rt}=Ake^{rt}
    y(1+Ae^{rt})=Ake^{rt}
    y=\frac{Ake^{rt}}{1+Ae^{rt}}
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  9. #9
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    Quote Originally Posted by topspin1617 View Post
    I don't know where this rt+C is coming from, but...

    \mathrm{ln}|y|-\mathrm{ln}|k-y|=rt+C
    \mathrm{ln}|\frac{y}{k-y}|=rt+C
    \frac{y}{k-y}=Ae^{rt} (if you want it like this, A=e^C)
    y=(k-y)Ae^{rt}
    y=Ake^{rt}-yAe^{rt}
    y+yAe^{rt}=Ake^{rt}
    y(1+Ae^{rt})=Ake^{rt}
    y=\frac{Ake^{rt}}{1+Ae^{rt}}
    Thanks very much. Basically I'm working through this question:

    Tricky integral-de.png

    I'm trying a step at a time!

    I'm finding it hard, nearly done A though now
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  10. #10
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    The next part says if y(0) = k/3 find y(t)

    So when t=0, y = k/3 ?

    k/3 = Ak/(1+A)

    k(1+A) = 3(Ak)

    k + Ak = 3Ak
    k = 2Ak
    2A = 1

    A= 1/2 . Would you agree with this?

    So y(t) = (1/2)ke^rt/ 1+ (1/2)e^rt
    Last edited by chr91; November 14th 2010 at 12:17 PM.
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  11. #11
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    Yes, and you can make it simpler if there is another part coming up.

    y(t) = \dfrac{\frac12 ke^{rt}}{1 + \frac12 e^{rt}}

    Multiply by 2/2;

    y(t) = \dfrac{ke^{rt}}{2 + e^{rt}}

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