Finding coefficients on a solved diff. equation

In summary, the problem is to find the steady state solution for an L-R-C circuit with specific values for L, R, and C, and an input voltage of 50cos(t). The solution involves using the equations L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0 and m^2 + 2m + 4 = 0 to find the particular solution qp(t), which is given by qp(t) = \frac{1}{29}(60cos(t) + 5sin(t)). Both the student and their book made minor errors in their calculations, but the student has confirmed that their solution is correct.
  • #1
Sparky_
227
5

Homework Statement



Find the steady state solution:
In an L-R-C circuit - L = 1, R = 2, C = 0.25, E(t) = 50cos(t)

Homework Equations





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
 
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  • #2
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
is ODE - ordinary differential equation?
(almost has to be based on usage)

-curious
 
  • #4
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
Actually both you and your book are wrong based on my checks. The answer should be

[tex]q_p(t) = \frac{1}{29}(60cos(t) + 5 sin(t))[/tex]
 
  • #6
thanks Mindscrape.

I did try my solution:

[tex]qp(t) = \frac {150}{13}cos(t) + \frac {100}{13}sin(t) [/tex]

into the ODE

and it works


[tex]qp'(t) = -\frac {150}{13}sin(t) + \frac {100}{13}cos(t) [/tex]

[tex]qp''(t) = -\frac {150}{13}cos(t) - \frac {100}{13}sin(t) [/tex]

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


[tex] -\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))) [/tex]

[tex] cos (t)(-\frac {150}{13} + \frac {200}{13} + \frac {600}{13}) + sin(t)(-\frac {100}{13} - -\frac {300}{13} + -\frac {400}{13})[/tex]

[tex] cos(t)(\frac {650}{13}) + 0 = 50cos(t)[/tex]
 
  • #7
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 [tex] (D^2+1)(D^2+2D+4) = 0 [/tex]


I bet the 25q caused you some grief?

Sorry
-Sparky
 
  • #8
Ah, okay, that makes sense then.
 

1. What is the purpose of finding coefficients on a solved differential equation?

The coefficients in a differential equation represent the constants or variables that affect the behavior of the equation. By finding these coefficients, we can better understand the relationship between the variables and make predictions about the behavior of the system.

2. How do you determine the coefficients in a differential equation?

To find the coefficients in a differential equation, we can use various methods such as separation of variables, substitution, or integration. The specific method used depends on the type of differential equation and the given initial conditions.

3. Can the coefficients in a differential equation change over time?

Yes, the coefficients in a differential equation can change over time if the system is dynamic and affected by external factors. In such cases, the coefficients may be represented as functions of time rather than constants.

4. Is it important to find the coefficients accurately in a differential equation?

Yes, finding the coefficients accurately is crucial in solving a differential equation. Even small errors in the coefficients can lead to significant differences in the predicted behavior of the system. Therefore, it is essential to use appropriate methods and techniques to determine the coefficients as accurately as possible.

5. How do the coefficients affect the solution of a differential equation?

The coefficients in a differential equation directly affect the form and behavior of the solution. For example, changing the value of a coefficient may result in a different type of solution, such as oscillatory or exponential, or may change the stability of the system. Therefore, understanding the coefficients is crucial in analyzing and predicting the behavior of a system described by a differential equation.

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