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**1. Homework Statement**

Find the steady state solution:

In an L-R-C circuit - L = 1, R = 2, C = 0.25, E(t) = 50cos(t)

**2. Homework 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