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Just solving 2 simultaneous differential equations - am i being thick!? watch

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    Ok, am I just being stupid?

    Aim is to find x(t). alpha, g, M and m are all constants.

    I've tried just solving it as a standard second order ODE but can't really get much further than the complementary function.
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    I'm not convinced of some of your manipulation; it seems you've written \sin \alpha \cos \alpha = \sin \frac{\alpha}{2}.

    But aside from that, find the complementary function, the extra term on the RHS is just a constant, so look for a PI of the form x = K for some constant K.
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    Sorry, i meant sin(2alpha)/2. Will have a go now..
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    (Original post by 3thr3e)
    Sorry, i meant sin(2alpha)/2. Will have a go now..
    Yes. After the correction integrate the equation twice with respect to x.
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    (Original post by DFranklin)
    I'm not convinced of some of your manipulation; it seems you've written \sin \alpha \cos \alpha = \sin \frac{\alpha}{2}.

    But aside from that, find the complementary function, the extra term on the RHS is just a constant, so look for a PI of the form x = K for some constant K.
    IGNORE. I'm BEING THICK as suspected!
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    (Original post by imzir)
    Yes. After the correction integrate the equation twice with respect to x.
    You can't just do that though can you. x is a function of t..
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    (Original post by 3thr3e)
    You can't just do that though can you. x is a function of t..
    sorry i meant integrate x with respect to t. sorry I am an alevel student.
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    (Original post by imzir)
    sorry i meant integrate x with respect to t.
    that's like saying what's the integral of y with respect to x and not knowing what the function y is!
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    (Original post by 3thr3e)
    that's like saying what's the integral of y with respect to x and not knowing what the function y is!
    ive done second order differential equations before but this looks wierd for some reason.

    Whats your particular integral?
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    (Original post by DFranklin)
    I'm not convinced of some of your manipulation; it seems you've written \sin \alpha \cos \alpha = \sin \frac{\alpha}{2}.

    But aside from that, find the complementary function, the extra term on the RHS is just a constant, so look for a PI of the form x = K for some constant K.
    Ok I think i've done it. Please could you check it?
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    The broad strokes are there, but you also need to consider the particular integral to make your solution complete.

    I have appended the detailed workings below for your reference, hope it helps. Peace.

 
 
 
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