Since the coefficients are constant, $$\ddot \xi + r\dot \xi -\omega_0^2 \xi=0 \hspace{1 cm }(1)$$ will have a solution of the form $\xi = e^{\lambda t}$, this will give you $$ \lambda^2 + r \lambda - \omega_0^2 = 0 \hspace{1 cm }(2)$$ The roots of equation of $(2)$ are $$ \lambda = \frac{-r \pm \sqrt{r^2 + 4 \omega_0^2}}{2} \hspace{1 cm }(3) $$ which gives you the solution of the form $$\xi = c_1 e^{\lambda_1} + c_2 e^{\lambda_2} \hspace{1 cm }(4)$$ since $\lambda$'s are real, it isn't oscillating which is not character of Harmonic Oscillator.
Depending on weather one of $\lambda$ is positive or both negative, the solution diverges exponentially or decays to rest.