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    (Original post by ViralRiver)
    Oh wow... using that reduces to 2\lambda =3t. Very weird oO. If anyone has an explanation I'd love to know .
    Can you see how this suggests that lambda is actually a polynomial of degree 1 in t for this choice of PI. I'm guessing this suggests that the particular integral for the whole DE is probably \frac{3}{2}t^3e^{2t} (since lambda is \frac{3}{2}t if your PI is chosen to be \lambda t^2e^{2t}).

    Why that's happened is a different question all together. One I don't have an answer to. I will check this out with my maths teacher tomorrow.
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    (Original post by Farhan.Hanif93)
    Can you see how this suggests that lambda is actually a polynomial of degree 1 in t for this choice of PI. I'm guessing this suggests that the particular integral for the whole DE is probably \frac{3}{2}t^3e^{2t} (since lambda is \frac{3}{2}t if your PI is chosen to be \lambda t^2e^{2t}).

    Why that's happened is a different question all together. One I don't have an answer to. I will check this out with my maths teacher tomorrow.
    Ahhh yes I see what you mean. I had further maths today so was able to ask my teacher but he was clueless xD - any luck?
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    You can write x=\lambda t^n e^{2t}, work it all through and then choose n at the end. It works fine and only takes a couple of minutes. I just checked.
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    (Original post by Get me off the £\?%!^@ computer)
    You can write x=\lambda t^n e^{2t}, work it all through and then choose n at the end. It works fine and only takes a couple of minutes. I just checked.
    I mean, it's clear than n=2 doesn't quite work but I think the bigger question was why not (That's what I'm interested in, anyway).
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    It's a conspiracy of coefficients.
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    (Original post by ViralRiver)
    Ok thanks .

    Something's popped up now which doesn't really make sense either =\ .

    \frac{d^2x}{dt^2}-4\frac{dx}{dt}+4x=3te^{2t}

    Now, for the particular integral, I would use \lambda t^2e^{2t} as that does not appear in the CF. However, the solution requires you to use \lambda t^3e^{2t} oO. Seems like I really don't understand it after all :< .
    Basically, each root in the AE that 'matches' the form of the PI will "lose" you a coefficient of t when you apply the DE to your guess for the PI.

    Since you have a repeated root, you're going to lose 2 powers of t, so that t^2e^(2t) will become e^2t (times a constant) and t^3 e^(2t) will become te^2t.

    It's perhaps worth pointing out that "... does not appear in the CF" doesn't by itself imply very much about whether or not it will work as a guess for the PI. The form of the RHS matters too.

    If you had \frac{d^2x}{dt^2}-4\frac{dx}{dt}+4x=3t^{99}e^{2t} then obviously you wouldn't expect \lambda t^2e^{2t} to work, even though it still doesn't appear in the CF.
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    I see that DFranklin has posted while I've been struggling with a dodgy connection, LaTeX and maths, probably making my post redundant but it took too long to just dump it so here it is.


    Suppose
    \frac{d^2x}{dt^2}-2\alpha\frac{dx}{dt} + \alpha ^2 x=At^m e^{\alpha t}

    and suppose x=\lambda e^{\alpha t} t^n

    then \frac{dx}{dt}=\lambda e^{\alpha t}(\alpha t^n + n t^{n-1})

    and \frac{d^2 x}{dt^2}=\lambda e^{\alpha t}(\alpha^2 t^n + 2\alpha t^{n-1} +n(n-1) t^{n-2})

    Substituting into the differential equation:


    \lambda e^{\alpha t}[\alpha^2 t^n + 2\alpha t^{n-1} +n(n-1) t^{n-2}]+


    \lambda e^{\alpha t}[-2 \alpha ^2 t^n  -2\alpha n t^{n-1})]+

    \lambda e^{\alpha t}[\alpha  ^2 t^n]

    the only term remaining after simplifying is \lambda e^{\alpha t} n(n-1) t^{n-2}
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