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anonymous

  • 5 years ago

Solve y'' + 4y = 1 by using Variation of Parameters.

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  1. amistre64
    • 5 years ago
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    tahts all the way in the back of the bok :)

  2. anonymous
    • 5 years ago
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    yeah, 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.

  3. amistre64
    • 5 years ago
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    i cant even begin to understand the notation in my textbook :) something about homogenuous and nonhomogenous stuff, and Wronksian coeefs, and Cramers rule...

  4. anonymous
    • 5 years ago
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    y'' + 4y = 1 r^2 + r = 0 does that help ?

  5. anonymous
    • 5 years ago
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    @amistre64: haha yeah, I think I did the Wronskians and the integration correctly though..

  6. anonymous
    • 5 years ago
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    @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.

  7. anonymous
    • 5 years ago
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    ill post my solution in a bit, please wait a sec.

  8. anonymous
    • 5 years ago
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    ok, let me see what yu got so far

  9. anonymous
    • 5 years ago
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    here you go: http://dl.dropbox.com/u/17638088/varparamprob.pdf

  10. anonymous
    • 5 years ago
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    so all yu wanna do is simplify that?

  11. anonymous
    • 5 years ago
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    i simplify the last two terms and I get (1/2)cos^2 x instead of getting 1/4 haha

  12. anonymous
    • 5 years ago
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    are yu sure you Yp is right?

  13. anonymous
    • 5 years ago
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    I 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..

  14. anonymous
    • 5 years ago
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    here'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

  15. anonymous
    • 5 years ago
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    hmm 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?

  16. anonymous
    • 5 years ago
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    \[\int\limits_{}^{}-{1 \over 2} \sin(2x) = {\cos(2x) \over 4 }\] shouldn't be that?

  17. anonymous
    • 5 years ago
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    oops 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

  18. anonymous
    • 5 years ago
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    wtf.. i believe that's wrong O.o On my Ti-89 i got different on so did i on this website: http://www.numberempire.com/integralcalculator.php

  19. anonymous
    • 5 years ago
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    lol, let's check if your answer gets me down the road to 1/4 though. :)))

  20. anonymous
    • 5 years ago
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    hehe ok

  21. anonymous
    • 5 years ago
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    lol, you're right, I got 1/4 in the end using your answer. =)))) thanks!!

  22. anonymous
    • 5 years ago
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    haha yes, i got 1/4 too. the sin^2(2x) + cos^2(2x) = 1 so (1/4)*1 = 1/4 hehe

  23. anonymous
    • 5 years ago
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    oh 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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