# Solution to the second order differential equation of $LCR$ series circuit.

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.

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

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

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

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.