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C4 separating variables help please

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    Hi! So the c4 exam is tomorrow, I am fairly confident about most of the topics integration and vectors etc and I can solve some differential equations but there are so many ways of separating them! In some past papers they separate the function and the number if there is one and sometimes they make it all a function and integrate 1 on the right hand side?!! WHY!
    The formula in the edexcel textbook isnt really applicable ... it says to integrate 1/g(y) dy = integral of f(x) dx but this isnt how mark schemes do it.
    Can somebody please explain the correct way of doing this? )':
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    Anybody?! It's really confusing me. ):
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    If \dfrac{dy}{dx} = f(x)g(y)

    then the only way to do it is

    \int \dfrac{1}{g(y)} dy = \int f(x) dx


    I think you must be misunderstanding the MS if you think they are doing anything else

    Do you have an example
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    (Original post by TenOfThem)
    If \dfrac{dy}{dx} = f(x)g(y)

    then the only way to do it is

    \int \dfrac{1}{g(y)} dy = \int f(x) dx


    I think you must be misunderstanding the MS if you think they are doing anything else

    Do you have an example

    (Original post by Baljit-Padda)
    Hi! So the c4 exam is tomorrow, I am fairly confident about most of the topics integration and vectors etc and I can solve some differential equations but there are so many ways of separating them! In some past papers they separate the function and the number if there is one and sometimes they make it all a function and integrate 1 on the right hand side?!! WHY!
    The formula in the edexcel textbook isnt really applicable ... it says to integrate 1/g(y) dy = integral of f(x) dx but this isnt how mark schemes do it.
    Can somebody please explain the correct way of doing this? )':
    I know what he's talking about, I think - when you separate the variables it's not always obvious what to keep on either side. e.g. dy/dx = 3x + 4, do you keep the 4 on the left or right?
    The answer, I'd say, is to look at what you're trying to get to and that will help you.

    The principle is the same as with trig identities - there're loads of ways of doing it, but only one will lead you to the answer.
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    (Original post by TenOfThem)
    If \dfrac{dy}{dx} = f(x)g(y)

    then the only way to do it is

    \int \dfrac{1}{g(y)} dy = \int f(x) dx


    I think you must be misunderstanding the MS if you think they are doing anything else

    Do you have an example


    Well yes I understand that method and I use it too but it gives me the wrong answer. For example theres the june 2010 question on this topic. 75dh/dt = 4-5h and so dh/dt = (4-5h)/75 Now in this particular example they use the whole thing together as our g(y) but what I'm stuck on is how they identify the g(y) and f(x).
    For example I would have thought the (4-5h) is g(y) and f(x) was 1/75 ):
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    (Original post by Junaid96)
    I know what he's talking about, I think - when you separate the variables it's not always obvious what to keep on either side. e.g. dy/dx = 3x + 4, do you keep the 4 on the left or right?
    The answer, I'd say, is to look at what you're trying to get to and that will help you.

    The principle is the same as with trig identities - there're loads of ways of doing it, but only one will lead you to the answer.
    \dfrac{dy}{dx} = 3x + 4

    gives

    y = \int 3x+4 dx


    there are no variables to separate

    You have to have f(x) x g(y) to separate
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    (Original post by Baljit-Padda)
    Well yes I understand that method and I use it too but it gives me the wrong answer. For example theres the june 2010 question on this topic. 75dh/dt = 4-5h and so dh/dt = (4-5h)/75 Now in this particular example they use the whole thing together as our g(y) but what I'm stuck on is how they identify the g(y) and f(x).
    For example I would have thought the (4-5h) is g(y) and f(x) was 1/75 ):
    It can be either

    constant multipliers can go on either side
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    but they integrate one over (4-5h)/75 so that they are integrating 75/(4-5h) on the left hand side and then they integrate 1 dt on the right hand side. Why is it not integrating 1/(4-5h) dh = integral 1/75 dt ?
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    OHH well I just tried both methods and get the same answer.... So will I still get maximum marks even if my working is not exactly what is written in the markscheme?
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    (Original post by Baljit-Padda)
    but they integrate one over (4-5h)/75 so that they are integrating 75/(4-5h) on the left hand side and then they integrate 1 dt on the right hand side. Why is it not integrating 1/(4-5h) dh = integral 1/75 dt ?
    It can be either

    \dfrac{dy}{dx} = 3xy

    works with

    \int \dfrac{1}{y} dy = \int 3x dx

    but also works with

    \int \dfrac{1}{3y} dy = \int x dx

    the 3 can go either side

    Sometimes it is easier to have it one side or the other but it will work either way
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    (Original post by Baljit-Padda)
    OHH well I just tried both methods and get the same answer.... So will I still get maximum marks even if my working is not exactly what is written in the markscheme?
    yes
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    Aw Okay Thank you so much! I'm very grateful!
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    np

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    (Original post by TenOfThem)
    np

    ^ don
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    does anyone have 17 january 2012 mark scheme

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Updated: June 20, 2012
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