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    Ok im having trouble with some of these questions,
    In question 2, im having trouble with part ii) i can find the auxillary equation and then find the general solution but it is not what the answer is, and i have no idea how to do part iv) in this question.
    Also in question 3, i've managed to work my way through part i) and ii) but i have no idea about the rest of the questions, any tips or a worked method would be very helpful thank you
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    (2)
    (ii)
    General solution for x:
    x = 2e^(-2t) + Ae^(-t) + Be^(2t)

    It follows from (1) that

    y
    = (1/18)(11x + e^(-2t) - x')
    = (1/18)[22e^(-2t) + 11Ae^(-t) + 11Be^(2t) + e^(-2t) - (-4e^(-2t) - Ae^(-t) + 2Be^(2t))]
    = (1/18)[27e^(-2t) + 12Ae^(-t) + 9Be^(2t)]
    = (3/2)e^(-2t) + (2/3)Ae^(-t) + (1/2)Be^(2t)

    (iii)
    Suppose that x -> 0 as t -> infinity. Then B = 0. So y -> 0 as t -> infinity.

    Suppose that y -> 0 as t -> infinity. Then B = 0. So x -> 0 as t -> infinity.

    (iv)
    Assume that the solutions do not tend to 0. Then B \neq 0.

    As t -> infinity,

    x ~ Be^(2t)
    y ~ (1/2)Be^(2t)
    y/x -> 1/2

    --

    Assume that the solutions do tend to 0. Then B = 0.

    There are two cases to consider: (a) A = 0; (b) A \neq 0.

    In Case (a), x = 2e^(-2t) and y = (3/2)e^(-2t). So y/x -> 3/4 as t -> infinity.

    In Case (b), as t -> infinity,

    x ~ Ae^(-t)
    y ~ (2/3)Ae^(-t)
    y/x -> 2/3

    So the possible limits are 3/4 and 2/3.
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    (Original post by geekypoo)
    Also in question 3, i've managed to work my way through part i) and ii) but i have no idea about the rest of the questions, any tips or a worked method would be very helpful thank you
    See attached.

    Edit: messed up this q a bit. Got it fixed now (hopefully !!)
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    Thanks alot, Fermat what does the top line of the iii) part say? that might be the bit that im missing. Because i still dont understand it (sorry for not being able to read it)
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    Which module would this be (Edexcel)?
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    Sorry about the illegibility

    terminal velocity is Vt = 70 m/s
    parachutisats velocity is v = 0.99*Vt
    or
    v/Vt = 0.99
    v/70 = 0.99
    =========
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    okies, thank you Fermat. it was just that one line the rest was fine.

    And this is OCR (MEI) syllabus, Mechanics 4 (a.k.a Differential Equations)
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    right
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    (Original post by geekypoo)
    okies, thank you Fermat. it was just that one line the rest was fine.

    And this is OCR (MEI) syllabus, Mechanics 4 (a.k.a Differential Equations)
    you can get the ocr past papers and MS/(examiner's reports) from here
    passsword is: sirisaac
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    I know i can get the mark scheme and stuff, but i sometimes cant really see where some of the steps have some from, but when i see a fully worked version i can see where i have made mistakes or where i haven't done something simple.
    I'll have another question posted if i cant get the worked answers from somewhere. But this time M3
    By the way Fermat, in part iv) in the mark scheme it says that B= -59.5 not 69.4 like it says in your answer, is that just error carried forward throughout parts iv and v then?
 
 
 
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