i am trying to figure out how to use the solutions of characteristic equations to find the frequencies of their related oscillators.
my first differential equation is this:
$y''[t] + 4 \pi \cdot y[t] = 0$
and the solutions to its characteristic equation are:
$z = +/-2\pi \cdot i$
i also have another one:
$20.1 \cdot y[t] + 5.7 \cdot y'[t] + y''[t] = 0$
its solutions to its characteristic equation are:
$-2.85 +/- 3.46085 \cdot i$
i have searched a lot online but i can't find any straightforward explanations for how to approach these problems. any suggestions? thanks in advance.
This is straightforward application of the method of the characteristic equation for solving linear, constant coefficient, ordinary differential equations. Having substituted $y = \mathrm{e}^{rt}$ into the equation and factoring out and cancelling the never zero $\mathrm{e}^{rt}$ factor, what remains is the characteristic polynomial. It's roots are the choices of $r$ that make the substitution work. So the solutions are $y(t) = \mathrm{e}^{r_1 t}$ and $y(t) = \mathrm{e}^{r_2 t}$. Now use Euler's formula to see that the real part of $r$ tells you about the amplitude (increasing with time if $\Re r > 0$ and decreasing with time if $\Re r < 0$) and the imaginary part tells you the frequency of oscillation, that frequency being $\frac{2\pi}{\Im r}$.
(Notation: $\Re$ is the real part of a complex number and $\Im$ is the imaginary part. So $\Re(1+2\mathrm{i}) = 1$ and $\Im(1+2 \mathrm{i}) = 2$.)
For your second example: \begin{align*} y''(t) + \frac{57}{10} y'(t) + \frac{201}{10} y(t) &= 0 \\ y(t) &= \mathrm{e}^{rt} \\ y'(t) &= r \mathrm{e}^{rt} \\ y''(t) &= r^2 \mathrm{e}^{rt} \\ r^2 \mathrm{e}^{rt} + \frac{57}{10} r \mathrm{e}^{rt} + \frac{201}{10} \mathrm{e}^{rt} &= 0 \\ (r^2 + \frac{57}{10} r + \frac{201}{10}) \mathrm{e}^{rt} &= 0 \\ r^2 + \frac{57}{10} r + \frac{201}{10} &= 0 \\ r &= \frac{1}{20}(-57 \pm \mathrm{i}\,\sqrt{4791}) \\ y_1(t) &= \mathrm{e}^{\frac{1}{20}(-57 + \mathrm{i}\,\sqrt{4791})t} \\ &= \mathrm{e}^{\frac{-57}{20}t} \mathrm{e}^{\frac{\sqrt{4791}}{20}\mathrm{i}t} \\ &= \mathrm{e}^{\frac{-57}{20}t} \left( \cos \left(\frac{\sqrt{4791}}{20}t \right) + \mathrm{i} \sin \left( \frac{\sqrt{4791}}{20}t \right) \right) \text{and} \\ y_2(t) &= \mathrm{e}^{\frac{1}{20}(-57 - \mathrm{i}\,\sqrt{4791})} \\ &= \mathrm{e}^{\frac{-57}{20}t} \mathrm{e}^{\frac{-\sqrt{4791}}{20}\mathrm{i}t} \\ &= \mathrm{e}^{\frac{-57}{20}t} \left( \cos \left(\frac{-\sqrt{4791}}{20}t \right) + \mathrm{i} \sin \left( \frac{-\sqrt{4791}}{20}t \right) \right) \end{align*}
Of course, the frequency of $\cos(at)$ is $\frac{2\pi}{a}$ and similarly for sine, obtaining the frequency $$ \frac{2\pi}{\frac{\sqrt{4791}}{20}} = 1.815{\dots} $$ with units that are the reciprocal of the units of $t$ (Hertz if $t$ is in seconds).