0
$\begingroup$

When applying Kirchoff's voltage law to a $\text{LCR}$ series circuit. The following differential equation pops up..

$$E\sin(Kt)-\dfrac{q}{C}-R\dfrac{dq}{dt}-L\dfrac{d^2q}{dt^2}=0$$

Where $E$, $K$, $R$, $C$ and $L$ are positive constants.

I tried solving the equation by assuming the solution to be of the form $A\sin(Bx)+C\cos(Dx)$ and then solving for the constants but it didn't work.

I'm in 12th grade and only know how to solve differential equations like linear first degree.

$\endgroup$
0
$\begingroup$

This is a second order ODE, inhomogeneous. You have to solve it by beginning with solving the homogeneous one

$$L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=0$$

passing by polynomial

$$L \lambda^2 + R\lambda +\frac{1}{C}=0$$

Try to solve it in $\lambda$ and you will obtain the costants that have to be inserted in the first draft of solution, that will be of the form

$$q(t) = c_1 e^{\lambda_1t} + c_2e^{\lambda_1t} \quad \text{with } \lambda_1\neq\lambda_2$$

$\endgroup$
  • $\begingroup$ How would you know that $\lambda_1 \ne \lambda_2$, it depends on $L,R,C$ and in this case, the homogeneous solution would be $q(t) = (At+B)e^{\lambda_1 t}$ $\endgroup$ – stity Sep 4 '17 at 10:15
  • $\begingroup$ If $\lambda_1 = \lambda_2$ then solution becomes of the form $$q(t)=c_1e^{\lambda t}+c_2 t e^{\lambda t}$$ $\endgroup$ – Clyde A. Jansen Sep 4 '17 at 10:17
  • $\begingroup$ And you don't explain how to find the whole solution (i.e. homogeneous + particular solution) $\endgroup$ – stity Sep 4 '17 at 10:19
  • $\begingroup$ Of course not. Just want to do it step by step, leading him to polyomial one... $\endgroup$ – Clyde A. Jansen Sep 4 '17 at 10:21
  • 1
    $\begingroup$ The particular solution is not a polynomial one but a trigonometric one of the form $A\cos(Kt)+B\cos(Kt)$ $\endgroup$ – stity Sep 4 '17 at 10:34
0
$\begingroup$

Don't know if you are acquainted with complex numbers.

These type of equations are easily solved (electrical engineers know very well) in the complex field.

Pass from $E\sin(Kt)$ to $V=E e^{iKt}$, and also express $q=Q e^{iK(t+\tau)}$, or better just $q=Q e^{iKt}$, allowing $Q$ to be complex .
To take the derivatives is easy, and you arrive to a complex equation to be solved for $Q$ and which is linear in $V$ ($V$ divided by a complex expression in $R,L,C$).

In that, take the immaginary component of both sides, and that's all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.