1
$\begingroup$

The RLC circuit forms the following differential equation:

$$\frac{di^2}{dt^2}+a_1\frac{di}{dt}+a_0i=k\cos{(wt+f)}$$

Assuming the roots of the characteristic equation are complex, complimentary function becomes:

$$i_c=e^{\alpha t}\left[ c_1\cos{\beta t} + c_2\sin{\beta t} \right] $$

So far, I tried "undetermined coefficients", "variation of parameters" and "solution by the use of complex variables" methods to find the particular solution $i_p$. However, things get messy for all of them and I can't find the particular solution.

Can you please show me the most practical way of finding the particular solution? Thank you.

$\endgroup$
7
  • $\begingroup$ Have you heard of Laplace transform? $\endgroup$
    – Troy Woo
    Sep 26, 2014 at 8:55
  • $\begingroup$ @TroyWoo There is no initial conditions. Additionaly, I don't want to use Laplace transform. It is also a circuit analysis method. I have been studying diff. equations for one month. I want to make use of it. $\endgroup$
    – user137035
    Sep 26, 2014 at 10:29
  • $\begingroup$ You will have to provide initial conditions...otherwise there is no method that can produce one for you... $\endgroup$
    – Troy Woo
    Sep 26, 2014 at 11:35
  • $\begingroup$ @TroyWoo I know. $\endgroup$
    – user137035
    Sep 26, 2014 at 11:40
  • $\begingroup$ Judging from what you said, you meant that the absence of initial conditions stops you from applying Laplace transformation, which is ridiculous. I think it would be helpful if you explain your logic here. $\endgroup$
    – Troy Woo
    Sep 26, 2014 at 11:53

1 Answer 1

1
$\begingroup$

I found the correct solution using "undetermined coefficients" method. It is more practical than "variation of parameters" and "solution by the use of complex variables" methods. Still this method doesn't look tempting. Here is what I found:

$$\boxed{i_p=A\cos(wt+f)+B\sin(wt+f)}$$

$$i^\prime_p=-wA\sin(wt+f)+wB\cos(wt+f)$$

$$i^{\prime \prime}_p=-w^2A\cos(wt+f)-w^2B\sin(wt+f)$$

Putting these into the equation:

$$-w^2A\cos(wt+f)-w^2B\sin(wt+f)-a_1wA\sin(wt+f)+a_1wB\cos(wt+f)+a_0A\\\cos(wt+f)+a_0B\sin(wt+f)=k\cos(wt+f)$$

$$\left[ (a_0-w^2)A+a_1wB \right]\cos(wt+f)+\left[ (a_0-w^2)B-a_1wA \right]\sin(wt+f)=k\cos(wt+f)$$

Finding the values of $A$ and $B$ is the most error-prone part. If you are not a mathematician and you are rusty on such operations you can struggle for 5 hours like me :). The correct reults are:

$$A=\frac{k(a_0-w^2)}{(a_0-w^2)^2+a_1^2w^2}\;\;and\;\;B=\frac{ka_1w}{(a_0-w^2)^2+a_1^2w^2}$$

$\endgroup$

You must log in to answer this question.