Diff Eq. - swapped coefficients help

[Sparky_]

Homework Statement

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

Find the steady state solution

Homework Equations

The Attempt at a Solution

L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 50cos(t)

\frac {di(t)}{dt} + 2 \frac {dq(t)}{dt} + 4q = 50cos(t)

q(t) = e^{-t}(c1cos(sqrt(12)t + c2sin(sqrt(12)t)

next annihilate 50*cos(t)
(D^2 +1) (D^2 + 2D + 4) = (D^2 +1)50cos(t)

roots = +/- i

qp(t) = Acos(t) + Bsin(t)

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

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

3A+2B = 50

3B-2A = 0

A = \frac {150}{13}

B = \frac {100}{13}

qp(t) = \frac {150}{13}cos(t) + \frac {100}{13}sin(t)

The book gets
qp(t) = \frac {100}{13}cos(t) - \frac {150}{13}sin(t)

I'm off with swapped coefficients and the sign.

Do I have something crossed?
Where is my error?

Thanks
-Sparky

 

[HallsofIvy]

Homework Statement

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

Find the steady state solution

Homework Equations

The Attempt at a Solution

L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 50cos(t)

\frac {di(t)}{dt} + 2 \frac {dq(t)}{dt} + 4q = 50cos(t)

Are we to assume that i= dq/dt?

q(t) = e^{-t}(c1cos(sqrt(12)t + c2sin(sqrt(12)t)

How did you get that? If you took i= dq/dt so that di/dt= d²q/dt², then characteristic equation is r²+ 2r+ 4= (r+2)²= 0 and the general solution (to the associated homogeneous equation) is
q(tt)= c1e^{-2t}+ c2 te^{-2t}

next annihilate 50*cos(t)
(D^2 +1) (D^2 + 2D + 4) = (D^2 +1)50cos(t)

roots = +/- i

qp(t) = Acos(t) + Bsin(t)

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

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

3A+2B = 50

3B-2A = 0

A = \frac {150}{13}

B = \frac {100}{13}

qp(t) = \frac {150}{13}cos(t) + \frac {100}{13}sin(t)

The book gets
qp(t) = \frac {100}{13}cos(t) - \frac {150}{13}sin(t)

I'm off with swapped coefficients and the sign.

Do I have something crossed?
Where is my error?

Thanks
-Sparky

 

[Sparky_]

You are correct on the first (homogeneous solution) - I had an obvious error in my work. I didn't recognize the obvious solution and made a mistake in factoring.

(However, I don't need that solution to get the steady state) I need the particular qp(t)

the (D^2 +1) (D^2 + 2D + 4) = (D^2 +1)50cos(t)

the second term is the solution to the homogeneous - as you pointed out q(tt)= c1e^{-2t}+ c2 te^{-2t}

The (D^2 +1)

roots = +/- i

with all the equating coefficients is where my error is.

Thoughts?

-Sparky