Oh, yeah. You do still have to use parts. 2 (or 1) is easy to integrate, at least. But starting with parts might actually work better. Or just as well.
Edit: Well, works okay, anyway. I was thinking that ln(x) would be easy to integrate and ln(x^2) wouldn't be integratable without changing it to 2ln(x) first.
edited 13th Jun '10 5:51:25 PM by Zizoz
This is a problem set in probability theory and statistics. I'm desperate and soliciting for help. If anyone wants to do this for the lulz, then go ahead. I'd appreciate any help I can get, since I'm really lost here (especially with what distribution to use for discrete random variables).
My FF.net accountDoes anyone know how to do a Decimal Quine-Mc Cluskey algorithm of boolean simplification for a function of five variables?
My FF.net accountI thought EEE 8 was circuit analysis for CS students? Also, I'm going to bed now, so nevermind. :p
My FF.net accountOh you smart smart guy. I checked back and got the prime implicants down, so I'm pretty close to solving this.
Thanks though. :p
My FF.net accountAh, christ. Can someone explain to me why yp here isn't just "2e^x", but some abomination borne out of being divided by a made-up variable D?
[1]
◊
Laplace?
See, this isn't my writing, this is the tutor's. THIS
◊ is what I did. As you can see, I have no idea what his notation is referring to.
- 1. For yc, yc^0 being the auxiliary formula and ycc being the r formula, r1 != r2, therefore yc = (C1)e^[(r1)x] (consulting my auxiliary roots result chart).
- 2. For yp, G(x) = Ae^(kx) (where k and A are constants), therefore yp = Ae^(kx). (consulting my g(x) chart.)
- y = yp + yc.
But evidently my yp is wrong. The tutor's work suggests that:
- yp is actually what I gave divided by the ycc, except with (d/dx) replacing the rs (and dx/d being an integral).
- For some reason, since if you consider G(x) to be e^ax in this case, a = D = 1, which can't be because then it's 1/0.
What am I missing here?
edited 4th Aug '10 4:46:55 PM by Barcode711
Worshipper of Ahura Mazda, as proclaimed by Zoroadster http://twitter.com/bpglobalprWhen he substituted in 1 for D it exploded because there was a zero on the bottom so he took the derivative of the top and bottom. So he wound up with 2x*exp(x) on top and the other stuff on the bottom. I'm not sure why he substituted 1 for D though. It's been too long since I've took calc. If they taught a better method afterward, I probably just remember that.
Fight smart, not fair.I made an accidental correction. It soud be if G(x) = Ce^kx, where C and k are constants, then yp = Ae^kx. I found the same tutor and we'll see if he can teach me how to solve for A...
Worshipper of Ahura Mazda, as proclaimed by Zoroadster http://twitter.com/bpglobalprYep, figured it out. It's a specific solving-for-the-constant method for these e^kx G(x) equations, which for some reason I don't have notes on. The others should be easy, as I have notes on them.
Thanks anyway.
Worshipper of Ahura Mazda, as proclaimed by Zoroadster http://twitter.com/bpglobalprI am disturbed by the fact that it took me about 5 minutes to understand either solution. D:
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I don't remember them and they aren't in my calculus book
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