# Finding coefficients on a solved diff. equation

1. Homework Statement

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

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

$$m^2 + 2m + 4 = 0$$ (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

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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?

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

-curious

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.

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

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

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

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

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

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

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

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

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

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

Ah, okay, that makes sense then.