I've been given the following problem:
For which $a\in\mathbb{R}$ will all nontrivial sol'ns of the eq $x''+ax'+4x=0$ be oscillatory about zero?
The characteristic exponents are
${r_{1,2}}^{*} = \frac{-a\pm\sqrt{a^2-16}}{2}$
and my book says that these sol'ns are oscillatory whenever $\Im(r^*)\neq0$.
How does this follow?
Define the variables
\begin{eqnarray} \frac{{\rm d}x}{{\rm d}t} &=& y \\ \frac{{\rm d}y}{{\rm d}t} &=& \frac{{\rm d}^2x}{{\rm d}t^2} = -a\frac{{\rm d}x}{{\rm d}t} - 4x = -ay - 4x \tag{1} \end{eqnarray}
In matrix form this is
$$ \frac{{\rm d}}{{\rm d}t} \begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -4 & -a \end{pmatrix} \begin{pmatrix}x \\ y \end{pmatrix} \tag{2} $$
or equivalently
$$ \frac{{\rm d}{\bf z}}{{\rm d}t} = A {\bf z} \tag{3} $$
The solution to this problem is
$$ {\bf z}(t) = e^{A t}{\bf z}_0 \tag{4} $$
So the problem is reduced to finding $e^{At}$, it $A$ is diagonalizable, then $A = U \Lambda U^{-1}$, where $\Lambda$ is diagonal matrix with the eigenvalues of $A$, remember that to find those we need to solve ${\rm det}(A - \lambda I) = 0$, which in this case is
$$ 0 = {\rm det}\begin{pmatrix} - \lambda & 1 \\ -4 & -a -\lambda\end{pmatrix} = \lambda(a + \lambda) + 4 ~~~\Rightarrow~~~ \lambda_{\pm} = \frac{-a \pm \sqrt{a^2 - 16}}{2} \tag{5} $$
(looks familiar?) Once you have this values, you just need to calculate the exponential $e^{A t}$, but that is easy since
\begin{eqnarray} e^{At} &=& \sum_{k=0}^{+\infty} \frac{(At)^k}{k!} = \sum_{k=0}^{+\infty} \frac{t^k}{k!}A^k \\ &=& \sum_{k=0}^{+\infty}\frac{t^k}{k!} (U\Lambda U^{-1})^k = \sum_{k=0}^{+\infty}\frac{t^k}{k!} U\Lambda^k U^{-1} \\ &=& U\left(\sum_{k=0}^{+\infty}\frac{t^k}{k!} \Lambda^k \right)U^{-1} = U\left(\sum_{k=0}^{+\infty}\frac{t^k}{k!} \begin{pmatrix} \lambda_1^k & 0 \\ 0 & \lambda_-^k \end{pmatrix}\right)U^{-1} \\ &=& U \begin{pmatrix} \sum_k t^k\lambda_1^k/k! & 0 \\ 0 & \sum_kt^k\lambda_-^k/k! \end{pmatrix} U^{-1} \\ &=& U \begin{pmatrix} e^{\lambda_+ t} & 0 \\ 0 & e^{\lambda_- t} \end{pmatrix} U^{-1} \tag{6} \end{eqnarray}
If you replace that into Eqn (4) you get the behavior of $x(t)$. What's important to realize here is that if $\lambda_{\pm}$ is complex, say $\lambda_{\pm} = -a/2 \pm i \omega_{\pm}$, then
$$ e^{\lambda_{\pm} t} = e^{(-a/2 \pm i \omega)t} = e^{-a/2}[\cos{\omega t} \pm i \sin(\omega t)] $$
which leads to an oscillatory solution!