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anonymous
 5 years ago
Solve y'' + 4y = 1 by using Variation of Parameters.
anonymous
 5 years ago
Solve y'' + 4y = 1 by using Variation of Parameters.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tahts all the way in the back of the bok :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, but i keep on getting this answer: http://dl.dropbox.com/u/17638088/myans.gif instead of getting just 1/4 as the last term.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i cant even begin to understand the notation in my textbook :) something about homogenuous and nonhomogenous stuff, and Wronksian coeefs, and Cramers rule...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y'' + 4y = 1 r^2 + r = 0 does that help ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@amistre64: haha yeah, I think I did the Wronskians and the integration correctly though..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@MathMind: that gives me the complementary solution y_c, which I have no problem getting. the particular solution y_p, on the other hand, is giving me trouble.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill post my solution in a bit, please wait a sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, let me see what yu got so far

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here you go: http://dl.dropbox.com/u/17638088/varparamprob.pdf

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so all yu wanna do is simplify that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i simplify the last two terms and I get (1/2)cos^2 x instead of getting 1/4 haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are yu sure you Yp is right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I'm sure.. but I put the original DE on WolframAlpha and it gives me 1/4 as the last term, so I posted this problem here to see if anyone can point out what's wrong with my answer..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here's WolframAlpha's take on the problem: http://www.wolframalpha.com/input/?i=y%27%27+%2B+4y+%3D+1 here's the simplification options of my answer: http://www.wolframalpha.com/input/?i=simplify+c_1*sin+2x+%2B+c_2*cos+2x+%2B+1%2F4+sin%5E2%282+x%29%2B1%2F2+cos%282+x%29+cos%5E2%28x%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm if that's what it is then you probablyt made a mistake somewhere i think your Yp is not right At U1, how did yu get the integral of (1/2)sin(2x) to be (1/2) cos^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}{1 \over 2} \sin(2x) = {\cos(2x) \over 4 }\] shouldn't be that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops that's a typo, that's supposed to be U_2. anyway, i put the integral on WolframAlpha just to be sure and it gave me this: http://www.wolframalpha.com/input/?i=integrate+%281%2F2%29sin+2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wtf.. i believe that's wrong O.o On my Ti89 i got different on so did i on this website: http://www.numberempire.com/integralcalculator.php

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, let's check if your answer gets me down the road to 1/4 though. :)))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, you're right, I got 1/4 in the end using your answer. =)))) thanks!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha yes, i got 1/4 too. the sin^2(2x) + cos^2(2x) = 1 so (1/4)*1 = 1/4 hehe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh and the reason that wolframalpha got that as a integral is because\[\cos(2x) =2 \cos^2(x) \]\[{ \cos(2x) \over 4 } = {2 \cos^2(x) \over 4 } = { \cos^2(x) \over 2 }\] but that doesn't help hehe
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