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    (Original post by oinkk)
    That's a weird one. For what its worth, I get \lambda = \frac{3}{2} with the particular integral  \lambda t^2e^2^t.

    I'd also be appreciative of anyone who could weigh in on this... since  \lambda t^2e^2^t doesn't form part of our complimentary function?
    hmm... yeah doing it with the P.I. they wanted gives  \lambda = 1/2

    I didn't realise it even gave a different answer


    Zacken ? have you got any words of wisdom?
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    (Original post by EmmaLouise759)
    I've seen them do this before as well. I think it's so that when you differentiate to get  dy/dx, and then when you differentiate again to get  d^2/dx^2, t is always in your differential value, (if you used  t^2, then by the time you work out  d^2y/dx^2, the t term would have disappeared).

    I don't see why it's necessary though, but I think that's their logic.
    In some other P.I.s the terms have disappeared by the first differential so that shouldn't matter should it? unless you are right and it does in this specific example
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    (Original post by oinkk)
    Just had a thought... I think it's to do with the fact that  \lambda te^2^t appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give  \lambda t^2e^2^t to account for it appearing in our complimentary function, then t again to give  \lambda t^3e^2^t to account for the repeated roots.

    Or that could be crazy talk.
    If you find out whether this is right or wrong, can you explain it to me in a little more detail please?
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    (Original post by Music With Rocks)
    Zacken ? have you got any words of wisdom?
    See below.

    (Original post by oinkk)
    That's a weird one. For what its worth, I get \lambda = \frac{3}{2} with the particular integral  \lambda t^2e^2^t.
    Check again, you don't get \lambda = \frac{3}{2}. You don't get any \lambda.

    (Original post by oinkk)
    Just had a thought... I think it's to do with the fact that  \lambda te^2^t appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give  \lambda t^2e^2^t to account for it appearing in our complimentary function, then t again to give  \lambda t^3e^2^t to account for the repeated roots.Or that could be crazy talk.
    Not so crazy, that's precisely it.
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    Do you guys think they'll ask us to draw polar curves?
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    (Original post by oinkk)
    Just had a thought... I think it's to do with the fact that  \lambda te^2^t appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give  \lambda t^2e^2^t to account for it appearing in our complimentary function, then t again to give  \lambda t^3e^2^t to account for the repeated roots.Or that could be crazy talk.
    (Original post by Zacken)
    Not so crazy, that's precisely it.
    (Original post by EmmaLouise759)
    If you find out whether this is right or wrong, can you explain it to me in a little more detail please?
    Yeah I am not sure I quite follow The CF surely takes into account the repeated roots so why do you need to allow for an extra t multiplier?

    (I know you are right Oinkk and Zacken, just trying to understand it haha)
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    (Original post by EmmaLouise759)
    If you find out whether this is right or wrong, can you explain it to me in a little more detail please?
    Edit: See post by Zacken below. Much better explanation.
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    (Original post by PhysicsIP2016)
    Yes that's right
    To find the value of x1, sub only the points where C meets the x-axis into the general equation to get (2-x1)^2 + (0-y1)^2 = r^2 and (-1-x1)^2 + (0-y1)^2 = r^2. If you then equate them, you do not need to find the value for r^2 and the y1^2 terms cancel out.
    Do a separate one using the points where C meets the y-axis in order to find the value of y1 and then x1 and y1 will be your centre coordinates.
    Finally got it! Thanks so much for your help, really appreciate it
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    So if it already appears in the CF, you multiply by t, but then if the CF has two equal roots, then you have to multiply by t again?
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    (Original post by oinkk)
    My logic is this:

    We have a complementary function with two constants to find, A and B. However, since our auxiliary equation only gives one solution, we cannot find the two constants as we need two independent solutions to solve for A and B.

    So in order to find these two constants, we multiply by t to give us our second independent solution to the general solution.
    But does it not technically give two solutions, they just happen to be equal?
     m^2-4m-1=0
     (m-2)(m-2)
     m=2

    EDIT: hmmm... does this always apply then when there are equal roots? you multiply by an extra term. I was assuming I had done examples like this before but looking back I don't think I have.
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    Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

    In general, if you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t} and your CF contains \alpha t e^{2t} and you have repeated roots, it is true that you need to use switch from using \alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    i.e: the default guess is \alpha e^{2t} and then you multiply by t when you find out that you have repeated roots and then multiply by t again when you find out that your CF contains \alpha t.

    In this case, you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}.

    Note the extra factor of t on the RHS.

    So your default guess is \alpha t e^{2t} since you want to match the form of the RHS.

    Then you find out your CF contains that term, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    Then you find out you have repeated roots, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}.
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    (Original post by somevirtualguy)
    So if it already appears in the CF, you multiply by t, but then if the CF has two equal roots, then you have to multiply by t again?
    See my post above.
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    (Original post by Zacken)
    Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

    In general, if you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t} and your CF contains \alpha t and you have repeated roots, it is true that you need to use switch from using \alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    i.e: the default guess is \alpha e^{2t} and then you multiply by t when you find out that you have repeated roots and then multiply by t again when you find out that your CF contains \alpha t.

    In this case, you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}.

    Note the extra factor of t on the RHS.

    So your default guess is \alpha t e^{2t} since you want to match the form of the RHS.

    Then you find out your CF contains that term, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    Then you find out you have repeated roots, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}.
    Better than I could ever explain it Cheers.
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    (Original post by Zacken)
    Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

    In general, if you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t} and your CF contains \alpha t and you have repeated roots, it is true that you need to use switch from using \alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    i.e: the default guess is \alpha e^{2t} and then you multiply by t when you find out that you have repeated roots and then multiply by t again when you find out that your CF contains \alpha t.

    In this case, you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}.

    Note the extra factor of t on the RHS.

    So your default guess is \alpha t e^{2t} since you want to match the form of the RHS.

    Then you find out your CF contains that term, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    Then you find out you have repeated roots, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}.
    Thank you for the clear explanation, will definitely make sure to remember this.

    I think that is what Oinkk was saying?
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    (Original post by Nerrad)
    Question: https://gyazo.com/3a02dd69a1250ddf22c5f7b51a4946aa
    Answer: https://gyazo.com/9b04261b67135f9f0cd792a1b961b03d
    2nd ODEs
    Basically I've used the transformation y=xv to make the DE go from y wrt. x to v wrt. x and found the GS of v wrt.x. When I sub the v back in since v y/x , which means everything on the RHS is to be times by x. I'm wondering why in the mark scheme only the (Asin3x+Bsin3x) is times by x but the (x^2/9)-(2/81) isn't? Is there a mistake?
    Just gonna repost this here cuz I don't think anyone saw it
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    (Original post by Nerrad)
    Just gonna repost this here cuz I don't think anyone saw it
    Yes, I would certainly multiply the whole thing by x. What paper is that from?
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    (Original post by Nerrad)
    Just gonna repost this here cuz I don't think anyone saw it
    Mistake in the markscheme.
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    (Original post by Zacken)
    Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

    In general, if you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t} and your CF contains \alpha t e^{2t} and you have repeated roots, it is true that you need to use switch from using \alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    i.e: the default guess is \alpha e^{2t} and then you multiply by t when you find out that you have repeated roots and then multiply by t again when you find out that your CF contains \alpha t.

    In this case, you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}.

    Note the extra factor of t on the RHS.

    So your default guess is \alpha t e^{2t} since you want to match the form of the RHS.

    Then you find out your CF contains that term, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    Then you find out you have repeated roots, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}.
    Thanks for this, just to check, is it the case that for any second order de with repeated roots that you multiply the PI by t or do you only do this if your second order de has repeated roots and contains one of the terms of the PI you were going to use? Thanks
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    Out of interest what was the original question, want to try this later now
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    (Original post by Zacken)
    Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

    In general, if you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t} and your CF contains \alpha t e^{2t} and you have repeated roots, it is true that you need to use switch from using \alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    i.e: the default guess is \alpha e^{2t} and then you multiply by t when you find out that you have repeated roots and then multiply by t again when you find out that your CF contains \alpha t.

    In this case, you have \lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}.

    Note the extra factor of t on the RHS.

    So your default guess is \alpha t e^{2t} since you want to match the form of the RHS.

    Then you find out your CF contains that term, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t}.

    Then you find out you have repeated roots, so multiply by t. \alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}.
    Is this always the case with repeated roots?
 
 
 
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