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    Hello everyone

    I'm a bit stuck on an FP2 second order differential equation - any help would be really appreciated!

    The equation I have to solve is:

    \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 10\sin t

    and the question says: 'Find the values of p and q so that  x = p\sin t + q\cos t is a solution of the differential equation.'

    I got the complementary function to be:

     x = e^{-t} (A\cos 2t + B\sin 2t) which I'm fairly sure is okay.

    The first thing I'm stuck on is whether to use the particular integral of:

     x = (\lambda \cos \omega t + \mu \sin \omega t)

    or to multiply that by t, because the form has already been used in the complementary function (sort of)... I don't really know what I'm talking about.

    I tried it both ways - with the extra t, I ended up with \lambda and \mu in terms of t, when in all the examples I've seen it should cancel out. Without the extra t, I got \lambda = -1 and \mu = 2.

    My next problem is getting the answer in the form  x = p\sin t + q\cos t. If what I have done so far is right, my general solution is:

     x = e^{-t} (A\cos 2t + B\sin 2t) - \cos t + 2\sin t

    ...which isn't in that form. (I've tried simplifying it down but it gets pretty messy because of the double angles and I'm not sure it would even work out.)

    Could someone point me in the right direction please?
    (Sorry if I'm being completely dim! :rolleyes:)
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    (Original post by Short_Round)
    ...
    What's the question actually asking you to do? Just find the values of p,q such that x=psint+qcost is a solution?
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    (Original post by ghostwalker)
    What's the question actually asking you to do? Just find the values of p,q such that x=psint+qcost is a solution?
    Yup
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    (Original post by Short_Round)
    Yup
    In which case, all you need do is substitute your value for x into the equation. Equate coefficents of cos and sine on either side to get two equations involving p and q (which I think you have done anyway), and solve for p,q, getting 2,-1 (which you got).

    End of!
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    (Original post by ghostwalker)
    In which case, all you need do is substitute your value for x into the equation. Equate coefficents of cos and sine on either side to get two equations involving p and q (which I think you have done anyway), and solve for p,q, getting 2,-1 (which you got).

    End of!
    Thanks for your help!

    So the question is just asking for the particular integral, not the full general solution?
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    (Original post by Short_Round)
    Thanks for your help!

    So the question is just asking for the particular integral, not the full general solution?
    It's not even asking for a solution. Just the values of p,q, such that....
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    (Original post by ghostwalker)
    It's not even asking for a solution. Just the values of p,q, such that....
    Okay... I'm a bit confused - so the values that I found for  \lambda and  \mu are the values of p and q? (sorry to be a pain!)

    I'm also a bit baffled that it's a ten mark question but it doesn't want a full solution. :confused:
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    (Original post by Short_Round)
    Okay... I'm a bit confused - so the values that I found for  \lambda and  \mu are the values of p and q? (sorry to be a pain!)
    Yes.

    I'm also a bit baffled that it's a ten mark question but it doesn't want a full solution. :confused:
    Can you post a photo/scan of the actual question then, or a link to it.
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    (Original post by ghostwalker)
    Can you post a photo/scan of the actual question then, or a link to it.
    I can't find it anywhere online (it's not an official Edexcel one) and I don't have any way of scanning / photographing it at the moment, sorry. I can give you the exact wording, though:

    5. \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 10\sin t

    (a) Find the values of p and q so that  x = p\sin t + q\cos t is a solution of the differential equation. (10 marks)


    There's a part b as well but I haven't got on to that yet, so I won't post it.
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    (Original post by Short_Round)
    Hello everyone

    I'm a bit stuck on an FP2 second order differential equation - any help would be really appreciated!

    The equation I have to solve is:

    \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 10\sin t

    and the question says: 'Find the values of p and q so that  x = p\sin t + q\cos t is a solution of the differential equation.'

    I got the complementary function to be:

     x = e^{-t} (A\cos 2t + B\sin 2t) which I'm fairly sure is okay.

    The first thing I'm stuck on is whether to use the particular integral of:

     x = (\lambda \cos \omega t + \mu \sin \omega t)

    or to multiply that by t, because the form has already been used in the complementary function (sort of)... I don't really know what I'm talking about.

    I tried it both ways - with the extra t, I ended up with \lambda and \mu in terms of t, when in all the examples I've seen it should cancel out. Without the extra t, I got \lambda = -1 and \mu = 2.

    My next problem is getting the answer in the form  x = p\sin t + q\cos t. If what I have done so far is right, my general solution is:

     x = e^{-t} (A\cos 2t + B\sin 2t) - \cos t + 2\sin t

    ...which isn't in that form. (I've tried simplifying it down but it gets pretty messy because of the double angles and I'm not sure it would even work out.)

    Could someone point me in the right direction please?
    (Sorry if I'm being completely dim! :rolleyes:)
     x = (\lambda \cos \omega t + \mu \sin \omega t) would be the one to use as your complementary function only contains real exponentials
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    (Original post by natninja)
     x = (\lambda \cos \omega t + \mu \sin \omega t) would be the one to use as your complementary function only contains real exponentials
    Thank you!

    I always get confused about whether to multiply through by t or not - in this case, would I only need to multiply by t if the particular integral was  \lambda e^{pt} ?
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    (Original post by Short_Round)
    Thank you!

    I always get confused about whether to multiply through by t or not - in this case, would I only need to multiply by t if the particular integral was  \lambda e^{pt} ?
    nope, you'd need to multiply by t if it was  \lambda e^{pt}cos(t) (or sin(t))
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    i need help with FP2 differential equations too, can someone come on skype/viber or something to work with me?
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    (Original post by natninja)
    nope, you'd need to multiply by t if it was  \lambda e^{pt}cos(t) (or sin(t))
    Ooh okay... so you multiply by t if the particular integral is in the exact same form as the complementary function, not just if the form is included in it?

    Thanks again!
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    (Original post by Short_Round)
    Ooh okay... so you multiply by t if the particular integral is in the exact same form as the complementary function, not just if the form is included in it?

    Thanks again!
    yeah, if that function of t is in the CF, otherwise you'll just get exactly the same solution :P
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    (Original post by natninja)
    yeah, if that function of t is in the CF, otherwise you'll just get exactly the same solution :P
    Woo hoo - I feel like a lightbulb has come on in my head!

    Thanks very much :bigsmile:
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    (Original post by Short_Round)
    Woo hoo - I feel like a lightbulb has come on in my head!

    Thanks very much :bigsmile:
    Awesome from now on ODEs are fun :P (no really they are quite simple once you get the hang of them :P)
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    (Original post by natninja)
    Awesome from now on ODEs are fun :P (no really they are quite simple once you get the hang of them :P)
    I'll take your word for it...

    I'm not too fond of substitutions! :erm:
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    (Original post by Short_Round)
    I'll take your word for it...

    I'm not too fond of substitutions! :erm:
    Neither :P especially in coupled ones... the joys of uni XD (I have a paper on just ODEs, complex numbers and wave mechanics...)
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    (Original post by natninja)
    Neither :P especially in coupled ones... the joys of uni XD (I have a paper on just ODEs, complex numbers and wave mechanics...)
    I'm so glad I'm not doing maths at uni! I did AS maths last year, then decided that doing AS further alongside A2 maths this year would be a good idea... all it's done is scare the bejesus out of me.
 
 
 
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