I have tried so hard to figure out the following questoin where I am asked to solve the differential equation given by $$I''+0.1I'+\frac{1}{c}I = sin(t)+4sin(5t)$$ where this is the circuit model for 2nd order differential equations.
I am asked to find the steady state solution but I am having a mess understanding the solutions for this problem.
Where does the $\tan(\phi(\omega))$ even come from here ? I can't wrap my head around this part.
On the other hand, How I did on the other hand was as follows where $$y_p = y_{p1} + y_{p2}$$ for the respective $\sin(t)$ and $\sin(5t)$ functions. Then what I did was for $$y_{p1} = (D^{2}+0.1D+\frac{1}{c})\bar{y} = e^{it}$$ where $\bar{y}$ is the imaginary part of the solution. Then I used the substitution rule for exponential input given by: $$y_{p1} = \frac{1}{(i)^(2)+0.1i+\frac{1}{c}}(\cos(t)+i\sin(t))$$ $$= \frac{1}{(\frac{1}{c}-1)+0.1i)}(\frac{(\frac{1}{c}-1)-0.1i)}{(\frac{1}{c}-1)-0.1i)}(\cos(t)+i\sin(t))$$ $$=\frac{(\frac{1}{c}-1)-0.1i}{(1-\frac{1}{c})^2+0.01}(\cos(t)+i\sin(t))$$ so that the particular solution to $\sin(t)$ is given by the equation above.
Solving for $y_{p2}$, it follows that I get: $$(D^2+0.1D+\frac{1}{c})I = 4\sin(5t)$$ $$D^2 + 0.1D + \frac{1}{c})\bar{I} = 4e^{(5i)t}$$ $$y_{p2} = \frac{4e^{(5i)t}}{(5i)^2+0.1(5i)+\frac{1}{c}}$$ $$ = \frac{4e^{(5i)t}}{(\frac{1}{c}-25)+0.5i} = \frac{4}{(\frac{1}{c}-25)+0.5i}(\frac{(\frac{1}{c}-25)-0.5i}{(\frac{1}{c}-25)-0.5i})(\cos(5t)+i\sin(5t))$$ $$\frac{4(\frac{1}{c}-25)-0.5i}{(25-\frac{1}{c})^2+0.25}(\cos(5t)+i\sin(5t))$$ so that $$y_{p_2} = \frac{4}{(\frac{1}{c}-25)^2+0.25}((\frac{1}{c}-25)\sin(5t)-0.5\cos(5t))$$
Lastly, combining the two particular equation yields: $$y_{p} = \frac{(\frac{1}{c}-1)-0.1i}{(1-\frac{1}{c})^2+0.01}(\cos(t)+i\sin(t)) + \frac{4}{(\frac{1}{c}-25)^2+0.25}((25-\frac{1}{c})\sin(5t)-0.5\cos(5t))$$
As a side note, the $C$ term is a constant that we are not explicityl told and I am not sure how Lastly, combining the two particular equation yields: $$y_{p} = \frac{(1-\frac{1}{c})-0.1i}{(1-\frac{1}{c})^2+0.01}(\cos(t)+i\sin(t)) + \frac{4}{(\frac{1}{c}-25)^2+0.25}((\frac{1}{c}-25)\sin(5t)-0.5\cos(5t))$$
As a side note, the $C$ term is a constant that we are not explicitly told. I compared the numerical values of my solution and the solution on matlab and the values are not the same, unfortunately. I am quite sure that my method is correct as I have used it for other problems as well, but have never seen the solution's method before. I would greatly appreciate any help as I have been thinking a long time about this.
That seems like a really messy presentation to me. Here's a simpler one:
$$LQ''+RQ'+EQ=e^{i \omega t}$$
where $E=1/C$. We guess that a particular solution is a (perhaps complex) multiple of $e^{i \omega t}$. Substituting that in and dividing through by $e^{i \omega t}$ gives
$$B(-\omega^2 L + i \omega R + E)=1.$$
Then $B=\frac{1}{-\omega^2 L + i \omega R + E}$, provided this denominator is not zero, so $Q=\frac{e^{i \omega t}}{-\omega^2 L + i \omega R + E}$. (When this denominator is zero, our guess at the start was wrong, and you have to do something else. However, this is impossible if $\omega$ and $R$ are both real and nonzero.)
You can write this in a more physically meaningful form by doing the complex number arithmetic to write $B=Ae^{i \phi}$. Then $A$ (a positive real number) is the amplitude of the solution and $\phi$ is related to the phase shift between the solution and the forcing.