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Finding coefficients on a solved diff. equation

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the steady state solution:
    In an L-R-C circuit - L = 1, R = 2, C = 0.25, E(t) = 50cos(t)

    2. Relevant equations



    3. The attempt at a solution
    [tex] L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0 [/tex]

    [tex] \frac {dq^2(t)}{dt^2} + 2 \frac {dq(t)}{dt} + \frac {1}{0.25}q = 50cos(t) [/tex]

    [tex] m^2 + 2m + 4 = 0 [/tex] (the homogeneous equation

    [tex] q(t) = e^(-t) ( c1*cos(sqrt(12)t) + c2*sin(sqrt(12)t) [/tex]

    annihilate 50cos(t)
    [tex] (D^2+1)(D^2+2D+4) = 0 [/tex]

    [tex] m1, 2 = +/- i [/tex]

    [tex]qp(t) = Acos(t) + Bsin(t) [/tex]

    [tex]qp'(t) = -Asin(t) + Bcos(t) [/tex]

    [tex]qp''(t) = -Acos(t) - Bsin(t) [/tex]

    plugging back into the eq.

    [tex] [-Acos(t) - Bsin(t)] -2Asin(t) + 2Bcos(t) + 4Acos(t) + 4Bsin(t) = 50 cos(t) [/tex]

    [tex] cos(t)[-A+2B + 4A] + sin(t)[-B -2A + 4B] = 50 cos(t) [/tex]

    [tex] 3A + 2B = 50 [/tex]

    [tex] 3B - 2A = 0 [/tex]

    [tex] A= \frac {150}{13} [/tex]

    [tex] B= \frac {100}{13} [/tex]

    [tex]qp(t) = \frac {150}{13}cos(t) + \frac {100}{13}sin(t) [/tex]

    the book has
    [tex] qp(t) = \frac {100}{13}cos(t) - \frac {150}{13}sin(t) [/tex]

    I differ on which coefficient goes where and I missed a sign.

    Do I have something crossed up / some careless error?

    Thanks
    -Sparky
     
  2. jcsd
  3. Apr 2, 2008 #2
    I can't see your error, and mathematica says the particular would be 60cos(t)/29. You can always check by plugging it back into the ODE. If it works then it works, right?
     
  4. Apr 2, 2008 #3
    is ODE - ordinary differential equation?
    (almost has to be based on usage)

    -curious
     
  5. Apr 3, 2008 #4
    Yep ODE is, I thought at least, a common terminology for ordinary differential equation, so you don't get PDEs confused with ODEs, and so you can describe separable solutions to PDEs as ODEs.
     
  6. Apr 3, 2008 #5
    Actually both you and your book are wrong based on my checks. The answer should be

    [tex]q_p(t) = \frac{1}{29}(60cos(t) + 5 sin(t))[/tex]
     
  7. Apr 3, 2008 #6
    thanks Mindscrape.

    I did try my solution:

    [tex]qp(t) = \frac {150}{13}cos(t) + \frac {100}{13}sin(t) [/tex]

    into the ODE

    and it works


    [tex]qp'(t) = -\frac {150}{13}sin(t) + \frac {100}{13}cos(t) [/tex]

    [tex]qp''(t) = -\frac {150}{13}cos(t) - \frac {100}{13}sin(t) [/tex]

    [tex] \frac {dq^2(t)}{dt^2} + 2 \frac {dq(t)}{dt} + 4q = 50cos(t) [/tex]


    [tex] -\frac {150}{13}cos(t) - \frac {100}{13}sin(t) + 2(-\frac {150}{13}sin(t) + \frac {100}{13}cos(t) + 4(\frac {150}{13}cos(t) + \frac {100}{13}sin(t))) [/tex]

    [tex] cos (t)(-\frac {150}{13} + \frac {200}{13} + \frac {600}{13}) + sin(t)(-\frac {100}{13} - -\frac {300}{13} + -\frac {400}{13})[/tex]

    [tex] cos(t)(\frac {650}{13}) + 0 = 50cos(t)[/tex]
     
  8. Apr 3, 2008 #7
    Mindscrape,

    I found a problem with my original posting - I had ... "25q" = 50 cos(t) ...

    it was 1/0.25 q or 4 q

    I did have the 4 in the [tex] (D^2+1)(D^2+2D+4) = 0 [/tex]


    I bet the 25q caused you some grief????

    Sorry
    -Sparky
     
  9. Apr 3, 2008 #8
    Ah, okay, that makes sense then.
     
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