Finding coefficients on a solved diff. equation

1. Apr 2, 2008

Sparky_

1. The problem statement, all variables and given/known data

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
$$L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0$$

$$\frac {dq^2(t)}{dt^2} + 2 \frac {dq(t)}{dt} + \frac {1}{0.25}q = 50cos(t)$$

$$m^2 + 2m + 4 = 0$$ (the homogeneous equation

$$q(t) = e^(-t) ( c1*cos(sqrt(12)t) + c2*sin(sqrt(12)t)$$

annihilate 50cos(t)
$$(D^2+1)(D^2+2D+4) = 0$$

$$m1, 2 = +/- i$$

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

$$qp'(t) = -Asin(t) + Bcos(t)$$

$$qp''(t) = -Acos(t) - Bsin(t)$$

plugging back into the eq.

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

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

$$3A + 2B = 50$$

$$3B - 2A = 0$$

$$A= \frac {150}{13}$$

$$B= \frac {100}{13}$$

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

the book has
$$qp(t) = \frac {100}{13}cos(t) - \frac {150}{13}sin(t)$$

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

Do I have something crossed up / some careless error?

Thanks
-Sparky

2. Apr 2, 2008

Mindscrape

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?

3. Apr 2, 2008

Sparky_

is ODE - ordinary differential equation?
(almost has to be based on usage)

-curious

4. Apr 3, 2008

Mindscrape

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.

5. Apr 3, 2008

Mindscrape

Actually both you and your book are wrong based on my checks. The answer should be

$$q_p(t) = \frac{1}{29}(60cos(t) + 5 sin(t))$$

6. Apr 3, 2008

Sparky_

thanks Mindscrape.

I did try my solution:

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

into the ODE

and it works

$$qp'(t) = -\frac {150}{13}sin(t) + \frac {100}{13}cos(t)$$

$$qp''(t) = -\frac {150}{13}cos(t) - \frac {100}{13}sin(t)$$

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

$$-\frac {150}{13}cos(t) - \frac {100}{13}sin(t) + 2(-\frac {150}{13}sin(t) + \frac {100}{13}cos(t) + 4(\frac {150}{13}cos(t) + \frac {100}{13}sin(t)))$$

$$cos (t)(-\frac {150}{13} + \frac {200}{13} + \frac {600}{13}) + sin(t)(-\frac {100}{13} - -\frac {300}{13} + -\frac {400}{13})$$

$$cos(t)(\frac {650}{13}) + 0 = 50cos(t)$$

7. Apr 3, 2008

Sparky_

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 $$(D^2+1)(D^2+2D+4) = 0$$

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

Sorry
-Sparky

8. Apr 3, 2008

Mindscrape

Ah, okay, that makes sense then.