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    Could someone please explain how 1/(y(y+1) integrates to ln(y/(y+1))

    Unless I've made a mistake leading upto this question then I can't see how it does...

    Thanks, Ryan.
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    (Original post by montyr)
    Could someone please explain how 1/(y(y+1) integrates to ln(y/(y+1))

    Unless I've made a mistake leading upto this question then I can't see how it does...

    Thanks, Ryan.
    Use partial fractions.

     \\ \frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}

    Remember also the laws of logarithms:

     \ln{a} - \ln{b} = \ln{\frac{a}{b}}
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    O right cheers! I did remember the laws of logs but and was trying to think how on earth that would get to lny-ln(y+1)

    THe whole partial fractions thing seems a hell of a lot of work for the number of marks!

    The whole question is solve the differential equation:

    sin^2(x+pi/6)dy/dx=2xy(y+1) and is worth 6 marks...
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    I don't see why you need partial fractions of any kind.

    Implicitly i get
    \displaystyle \sin^2(x+\pi/6)=\frac{x^2}{y+1}+\frac{A_{cons  tant}}{y}
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    Well thats not what they're asking for in the question

    And I don't see how you can solve it implicitly?

    We've been taught to get all the x's and the dx on one side and all the ys and the dy on the other...
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    (Original post by montyr)
    Well thats not what they're asking for in the question

    And I don't see how you can solve it implicitly?

    We've been taught to get all the x's and the dx on one side and all the ys and the dy on the other...
    Doesn't it just say solve this ODE?

    It's separable and the the variables are independent. So can't you do:
    

sin^2(f(x))\frac{dy}{dx}=2xy/(y+1)

    then \displaystyle \int sin^2(f(x))dy=\int 2xy/(y+1) dx+A

    giving

    \displaystyle ysin^2(f(x))=x^2 y/(y+1)+A\\

\Rightarrow \sin^2(x+\frac{\pi}{6})=\frac{x^  2}{y+1}+A/y

    Where f(x)=x+pi/6
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    (Original post by Daniel Freedman)
    Use partial fractions.

     \\ \frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}

    Remember also the laws of logarithms:

     \ln{a} - \ln{b} = \ln{\frac{a}{b}}

    would the same rule apply for [1/3(1 +2x)] ?

    i don't see why you can't multiply out the denominator to get 1/3+6x
    and integrate to get 1/6ln|3 + 6x| ???
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    (Original post by Goldenratio)
    \displaystyle \int sin^2(f(x))dy=\int 2xy/(y+1) dx+A

    giving

    \displaystyle ysin^2(f(x))=x^2 y/(y+1)+A
    Uh, only if x and y are independent, which they're clearly not, otherwise the whole concept of a differential equation in x and y wouldn't make much sense. This equation is separable, but you haven't bothered to separate it.
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    (Original post by sheena18)
    would the same rule apply for [1/3(1 +2x)] ?

    i don't see why you can't multiply out the denominator to get 1/3+6x
    and integrate to get 1/6ln|3 + 6x| ???
    You can. You can also get 1/6 ln|1 + 2x|. The trick lies in noticing that the difference between these two answers is just +1/6 ln 3, which is a constant and so can get added to (or subtracted from) the constant of integration as necessary. This is one case where you really need that +c.
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    generalabriety, i dont understand what you mean to be honest with you
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    (Original post by sheena18)
    generalabriety, i dont understand what you mean to be honest with you
    Don't worry, it doesn't really matter. The way you were doing it is correct. Generalebriety was just showing a different version of what is effectively the same result (just with a different constant of integration) - you would get the same answers if you were to substitute in limits.

     \int \frac{1}{3(1+2x)} dx = \int \frac{1}{6x+3} = \frac{1}{6}\ln{|6x+3|} + C

    or integrating slightly differently:

     \int  \frac{1}{3(1+2x)} dx = \frac{1}{3} \int \frac{1}{2x+1} dx = \frac{1}{3} \times \frac{1}{2}\ln{|2x+1|} + D = \frac{1}{6}\ln{|2x+1|} + D
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    (Original post by sheena18)
    generalabriety, i dont understand what you mean to be honest with you
    I assumed your question was "the answer given is 1/6 ln|1 + 2x|, but if you integrate it differently you get 1/6 ln|3 + 6x|, which is right?", to which my answer was "they're both the same thing, if you add an arbitrary constant onto the end".
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    thanks daniel

    and just to check, could we use your method (with the partial fractions) for this question?

    EDIT: thanks for clarifying for me generalabriety
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    (Original post by sheena18)

    and just to check, could we use your method (with the partial fractions) for this question?
    Not really, no. Only use partial fractions if you have two (or more) factors that contain x (or whatever variable you're integrating with respect to) in the denominator e.g.  \frac{2}{(3x-2)(x+1)} If you just have a constant and one factor in the denominator (as is the case above) it's just a straightforward ln integration.
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    ah i see, makes sense. thanks again daniel!!
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    i thought i'd just put my query here since it says C4 integration

    this si first example in the book ...so i t'sprobably something very basic i'm missing out on [strike]as always then [/]

    \int \frac{cosx}{sin^2x} -2e^x dx



    it goes \int \frac{cosx}{sinx} \times \frac{1}{sinx} dx is
    \int cotxcosecx dx
    up to there i'm ok
    but then...book gives -cosecx -2e^x

    how did it do that ? :confused:
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    (Original post by tami*)
    i thought i'd just put my query here since it says C4 integration

    this si first example in the book ...so i t'sprobably something very basic i'm missing out on [strike]as always then [/]

    \int \frac{cosx}{sinx} -2e^x dx



    it goes \int \frac{cosx}{sinx} \times \frac{1}{sinx} dx is
    \int cotxcosecx dx
    up to there i'm ok
    but then...book gives -cosecx -2e^x

    how did it do that ? :confused:
    ???

      \\ \displaystyle \int \frac{\cos{x}}{\sin{x}} - 2e^x dx \\

\\ = \int \cot{x} - 2e^x dx \\

\\ = \ln{|\sin{x}|} - 2e^x + C

    EDIT: Are you sure it isn't  \displaystyle \int \frac{\cos{x}}{\sin^2{x}} - 2e^x dx ?

    If so, can you see what they've done?
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    sorry...i edited!

    noo...i'm stuck on the bit after they did that
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    (Original post by Daniel Freedman)
    ???

      \\ \displaystyle \int \frac{\cos{x}}{\sin{x}} - 2e^x dx \\

\\ = \int \cot{x} - 2e^x dx \\

\\ = \ln{|\sin{x}|} - 2e^x + C
    if that were a question... why is it ln sin and not lnsec^2x?
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    (Original post by Goldenratio)
    Doesn't it just say solve this ODE?

    It's separable and the the variables are independent. So can't you do:
    

sin^2(f(x))\frac{dy}{dx}=2xy/(y+1)

    then \displaystyle \int sin^2(f(x))dy=\int 2xy/(y+1) dx+A

    giving

    \displaystyle ysin^2(f(x))=x^2 y/(y+1)+A\\

\Rightarrow \sin^2(x+\frac{\pi}{6})=\frac{x^  2}{y+1}+A/y

    Where f(x)=x+pi/6
    I'm not quite sure what you're doing there tbh!
 
 
 
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