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Diff Eq. - swapped coefficients help

  • Thread starter Sparky_
  • Start date
194
1
1. 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

2. Homework Equations



3. The Attempt at a Solution

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

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

[tex] q(t) = e^{-t}(c1cos(sqrt(12)t + c2sin(sqrt(12)t) [/tex]

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

[tex] roots = +/- i[/tex]

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

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

[tex] cos(t)[-A+2B+4A] + sin(t)[-B-2A+4B] = 50cos(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 gets
[tex] qp(t) = \frac {100}{13}cos(t) - \frac {150}{13}sin(t) [/tex]

I'm off with swapped coefficients and the sign.

Do I have something crossed?
Where is my error?

Thanks
-Sparky
 
Last edited:

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,738
899
1. 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

2. Homework Equations



3. The Attempt at a Solution

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

[tex] \frac {di(t)}{dt} + 2 \frac {dq(t)}{dt} + 4q = 50cos(t) [/tex]
Are we to assume that i= dq/dt?

[tex] q(t) = e^{-t}(c1cos(sqrt(12)t + c2sin(sqrt(12)t) [/tex]
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
[tex]q(tt)= c1e^{-2t}+ c2 te^{-2t}[/tex]

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

[tex] roots = +/- i[/tex]

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

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

[tex] cos(t)[-A+2B+4A] + sin(t)[-B-2A+4B] = 50cos(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 gets
[tex] qp(t) = \frac {100}{13}cos(t) - \frac {150}{13}sin(t) [/tex]

I'm off with swapped coefficients and the sign.

Do I have something crossed?
Where is my error?

Thanks
-Sparky
 
194
1
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 [tex] (D^2 +1) (D^2 + 2D + 4) = (D^2 +1)50cos(t) [/tex]

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

The [tex] (D^2 +1) [/tex]

[tex] roots = +/- i[/tex]

with all the equating coefficients is where my error is.

Thoughts?

-Sparky
 

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