Are all solutions to the ODE $ay''(t) + by'(t) + cy(t) = 0$ of the form $y(t)= \alpha e^{(\beta + i\gamma)t}$? Let $a$ $b$ and $c$ be complex numbers. Consider the complex solution of the ODE
$$ay''(t) + by'(t) + cy(t) = 0.$$
If there exist solutions to this, are they necessarily of the form
$$y(t)= \alpha e^{(\beta + i\gamma)t}$$
for some constants $\alpha, \beta, \gamma$? Here $i$ is the imaginary unit. 
In every example I've seen, this has been the case. But I am not sure if it necessarily the case. 
 A: The given ODE is a linear homogeneous equation of second order. This implies that the set of solutions is a two-dimensional complex vector space. 
If the equation $a\lambda^2+b\lambda +c=0$ has two different solutions $\lambda_1$, $\lambda_2\in{\mathbb C}$ then the general solution is given by
$$y(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}\ .$$
If the equation $a\lambda^2+b\lambda +c=0$ has one double solution $\lambda_0\in{\mathbb C}$ then the general solution is given by
$$y(t)=(C_0+C_1 t)e^{\lambda_0 t}\ ,$$
as explained in all textbooks. Your impression and reminiscences  from seen examples led you to  too simple expectations.
A: If $a = 0$ and $b = 0$, then we have $c \, y(t) = 0$, which is not even an ODE. If $c \neq 0$, then the solution is $y (t) = 0$. If $c = 0$, every function from $\mathbb R$ to $\mathbb C$ is a solution.
If $a = 0$ and $b \neq 0$, then we have a 1st order ODE
$$\dot y + \left(\frac{c}{b}\right) y = 0$$
whose solution is $y (t) = \beta \, \exp\left(- \dfrac{c}{b} \, t\right)$.
If $a \neq 0$, then we have a 2nd order ODE
$$\ddot y + \left(\frac{b}{a}\right) \dot y + \left(\frac{c}{a}\right) y = 0$$
Let $x_1 := y$ and $x_2 := \dot y$. Then, the 2nd order ODE can be rewritten in the form
$$\begin{bmatrix} \dot x_1\\ \dot x_2\end{bmatrix} = \begin{bmatrix} 0 & 1\\ -\frac{c}{a} & -\frac{b}{a}\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix}$$
If $a,b,c \in \mathbb R$, then the eigenvalues of the matrix above will be complex conjugate pairs, and the solutions to the 2nd order ODE will be of the form
$$\beta_1 \exp((\sigma + i \omega) t) + \beta_2 \exp((\sigma - i \omega) t)$$
If the matrix is not diagonalizable and eigenvalue $\sigma + i \omega$ has multiplicity $2$, then the solutions are of the form
$$\beta_1 \exp((\sigma + i \omega) t) + \beta_2 \, t  \exp((\sigma + i \omega) t)$$
If $a,b,c \in \mathbb C$, then the eigenvalues of the matrix above will, in general, not be complex conjugate pairs. Thus, the solutions to the 2nd order ODE will be of the form
$$\beta_1 \exp((\sigma_1 + i \omega_1) t) + \beta_2 \exp((\sigma_2 + i \omega_2) t)$$
where, in general, $\sigma_1 \neq \sigma_2$ and $\omega_2 \neq -\omega_1$. If the matrix is not diagonalizable and $\sigma_1 = \sigma_2$ and $\omega_2 = \omega_1$, then the solutions are of the form
$$\beta_1 \exp((\sigma_1 + i \omega_1) t) + \beta_2 \, t \exp((\sigma_1 + i \omega_1) t)$$
