I am trying to look a bit deeper into the mathematics the equation of motion used in physics and engineering. I have some specific questions at the end, but please correct me if I make a mistake in my statement.
I have an ordinary, second-order, linear, homogeneous differential equation:
$ M \ddot{y} + C \dot{y} + K y = 0 $
where $M$, $C$ and $K$ are real valued.
I am familiar with an Ad-hoc method of solving this equation. Namely, I assume a solution takes form of
$ y_k(t) = Y_k e^{\omega_k t} $
substitute the form into the ODE,
$ \left( M \omega_k^2 + C \omega_k + K\right) y_k = 0, $
and solve for $\omega_k$ such that $y_k \neq 0$.
If $y$ is in dimension $N$, we generally have $2N$ solutions, where for each $k$, $\omega_k$ and its complex conjugate $\tilde{\omega}_k$ are solutions. Since our ODE is linear, the general solution is a combination of the individual solutions
$ y(t) = \sum_{k = 1}^N Y_k e^{\omega_k t} + \tilde{Y}_k e^{\tilde{\omega}_k t} $
The values of $Y_k$ and $\tilde{Y}_k$ are determined from initial values, i.e. $y(0) = y_0$, and further restrictions common in physics, such as $y(t)$ must be real for all $t \geq 0$.
Now here are my questions:
1 - If we select the form $y_k(t) = Y_k e^{ \mathbf{i}\omega_k t}$, am I correct to say that the roots $\omega_k$ and $\tilde{\omega}_k$ are no longer complex conjugates, but $-1$ times their complex conjugates?
2 - How do we know that the general solution cannot contain terms which are not representable by the exponential form? In other words, is our general solution truly general?