2nd order linear differential equation 
Attempt:
a) oscillating solutions will occur for $\alpha^2 < 4\beta$, no oscillation if $\alpha^2 > 4\beta$ with $\beta > 0$. (is this necessary?)
for some $\sigma$, we have $(\lambda-\sigma)^2 = \lambda^2 -2\sigma \lambda + \sigma^2$. Thus, we will have repeated roots if $\alpha = -2\sigma$ and $\beta = \sigma^2$.  
b) Check here for the answer to the first half. For $\lim_{t \to \infty} y = \infty$, we have that $\alpha < 0$, as we need $\frac{b}{a} <0$, so that $-\frac{b}{2a}>0$.   
c) I believe for a) this is just the middle region of $\alpha < 2\sqrt{\beta}$ and $\alpha> -2\sqrt{\beta}$. for b), when $\alpha >0$, we have decay, when $\alpha<0$ we have growth. Is this true?  
d) Naturally this is true if we have damped oscillation, that is, if we have $\alpha^2 < 4\beta$, and $\alpha > 0$. Is this correct?  
Thank you. Any thoughts appreciated. Will be awarding bounty for good answers. Anywhere from 50-250 bounty will be awarded, for a quick answer. 
 A: You have that the characteristic equation is
$p(\lambda)=\lambda^2+\alpha\lambda+\beta=0$.
Then you have of course:
$\lambda_{1,2}=\dfrac{-\alpha\pm\sqrt{\alpha^2-4\beta}}{2}$.
So


*

*Complex roots $\iff$ $\alpha^2<4\beta$

*No oscillations $\iff$ $\alpha^2>4\beta$

*Repeated roots $\iff$ $\alpha^2=4\beta$


Furthermore for asymptotic stability you require $-\alpha\pm\sqrt{\alpha^2-4\beta}<0$, which is satisfied in the first quadrant of the plane $\alpha-\beta$. Recall that since you have a linear system, asymptotic stability means that the eigenvalues have negative real part. See below a somewhat crappy diagram showing some examples of behaviour. Hope it helps.

I show some examples of solutions. Of course the particular solutions depend on the initial conditions.
Inside the parabola you have complex roots. At the parabola you have repeated eigenvalues. Outside the parabola you have no oscillations. At $\alpha=0$, $\beta>0$ you have periodic orbits because the  eigenvalues are purely imaginary. For $\beta<0$ you always have unstable solutions as the equilibrium point is always a saddle.
I cannot see the video but I guess from the diagram it should be clear.
