So, in class we studied the rlc circuit with a force ac voltage across the circuit. The teacher assumed a lot of things and we ended up with a sinusoidal function for current with some equations to calculate the voltages across parts of the circuit.

However, I'm trying to solve the differential equation from scratch. So I decided to complexifying the forced AC voltage using Euler's identity and take only the real part of the solution when I finished. $$ L\ddot{y}+R\dot{y}+\frac{1}{C}y= E\cos(\omega t) $$ $$ L\ddot{y}+R\dot{y}+\frac{1}{C}y= E\ e^{\omega it} .$$

I managed to find the complimentary solution for the differential equation, however I'm having a hard time finding the particular solution.

So far I assumed that the complementary solution looks like this.

$$y_p = Ae^{\omega it} $$

And I got A as..

$$A = \frac{E}{(-\omega^2L+\frac{1}{C})+\omega Ri} $$

I multiplied A with 1 but in fraction form with both the numerator and the denominator as the complex conjugate of the denominator of A.

$$A = \frac{(-\omega^2L+\frac{1}{C})-\omega Ri}{(-\omega^2L+\frac{1}{C})-\omega Ri} \cdot \frac{E}{(-\omega^2L+\frac{1}{C})+\omega Ri} $$

From here I have trouble simplifying the equation to form a simple sinusoidal equation. So if anyone can show me how you simplify this particular solution I would appreciate it.

Also, I'm confused on how my complementary solution has a exponential decay function. Since I thought the circuit should always have current flowing even after a long time.


1 Answer 1


As you have already done, set $y_p = Ae^{i\omega t}$, and solve for $A$. Now that you have found $A$ to be complex, it would be easier to immediately write it as $A = re^{i\theta}$. Recall for a complex number $z = x + iy$, $|z|^2 = x^2 + y^2$ and $\theta = \arctan(y/x)$. This should help you simplify your particular solution since $A$ can be expressed as $A = \frac{E}{r}e^{-i\theta}$, so $y_p = \frac{E}{r}e^{i(\omega t - \theta) }$, and note that if you did it correct, you will see a resonance at $\omega^2 = \frac{1}{LC}$.

I do not believe your particular solution has an exponential decay. In fact, given the right frequency, your particular solution will dominate (at the resonance!).

It seems that you have not solved for your homogeneous (or complementary) solution as yet. The homogeneous equation has a resistor (or damping) term, which will provide the exponential decay. This will cause this solution to die away and only the particular solution will remain for long $t$.

  • $\begingroup$ Thank you for the explanation. However, I don't see how the complementary solution is not an exponential decay since there is a term e^(-R/2L)*t in front of a sinusoidal function. $\endgroup$
    – Nick Yarn
    Commented Apr 28, 2018 at 19:46
  • $\begingroup$ More over using www-thphys.physics.ox.ac.uk/people/JuliaYeomans/… the pdf states that the steady state solution is only the particular solution. The complementary solution decays after sometime since it is a tern used to described the initial conditions of the circuit when their was no current flowing. $\endgroup$
    – Nick Yarn
    Commented Apr 28, 2018 at 20:16
  • $\begingroup$ Yes, I'm used to using "homogeneous and particular" versus "complementary and particular". I edited the answer to reflect both. So as you said, the complementary solution has an exponential decay. $\endgroup$
    – bbachu
    Commented Apr 29, 2018 at 1:54

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