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# FP2 Second ODE Problem watch

1. (Original post by ViralRiver)
Oh wow... using that reduces to . 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 (since lambda is if your PI is chosen to be ).

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.
2. (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 (since lambda is if your PI is chosen to be ).

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?
3. You can write , work it all through and then choose n at the end. It works fine and only takes a couple of minutes. I just checked.
4. (Original post by Get me off the £\?%!^@ computer)
You can write , 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).
5. It's a conspiracy of coefficients.
6. (Original post by ViralRiver)
Ok thanks .

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

Now, for the particular integral, I would use as that does not appear in the CF. However, the solution requires you to use 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 then obviously you wouldn't expect to work, even though it still doesn't appear in the CF.
7. 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

and suppose

then

and

Substituting into the differential equation:

the only term remaining after simplifying is
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