Stability of $y''+4y'+4y=0$ Decide whether the solution is asymptotically stable, stable but not asymptotically stable, or unstable
$y''+4y'+4y=0, \phi(t)=e^{-2t}$
After computation of the eigenvalues I found that it has eigenvalues $\lambda=-2$ with multiplicity 2, nonetheless, we find that it has general solution 
$\phi(t) = c_{1}e^{-2t} + c_{2}te^{-2t}$, 
since both eigenvalues are negative real parts then by Theorem the zero solution should be asymptotically stable, however, when I try inputting the solution into wolfram alpha for a graph the vectors seem to run towards $+\infty$. Is it not asymptotically stable, or is it and am I just overthinking this question?
 A: Simply find the limit as $x \to +\infty$ of the general solution $y(t) = c_1e^{-2t} + c_2e^{-2t}t$ :
$$\lim_{t \to +\infty} y(t) = \lim_{t \to +\infty} (c_1e^{-2t} + c_2e^{-2t}t) = 0$$
I will leave up the elaborated calculation of the limit above to you, but it's easy to check that it's true since the exponential is the dominant term.
Which tells us that the solution is globally stable. This can easily be checked simply by the fact that the $r$ term of $e^{rt}$ is negative, since $r = -2 <0$.
Below, I will just attach a stability lemma-elaboration for you to keep in mind regarding such cases :

Consider the homogeneous linear second-order equation
  $$x''(t) + ax'(t) + bx(t) = 0.$$
  If $b ≠ 0$, this equation has a single equilibrium, namely $0$. (That is, the only constant function that is a solution is equal to $0$ for all $t$.) To study the stability of this equilibrium, consider separately the three possible forms of the general solution of the equation, as given in a previous result.
Characteristic equation has two real roots
  If the characteristic equation has two real roots, $r$ and $s$, the general solution of the equation is $Ae^{rt} + Be^{st}$. This function converges to $0$ for all values of $A$ and $B$, so that the equilibrium is globally stable, if and only if $r < 0$ and $s < 0$.
Characteristic equation has a single real root
  If the characteristic equation has a single real root, $r$, the general solution of the equation is $(A + Bt)e^{rt}$. This function converges to $0$ for all values of $A$ and $B$, so that the equilibrium is globally stable, if and only if $r < 0$. (If $r < 0$ then for any value of $k$, $t^ke^{rt}$ converges to $0$ as $t$ increases without bound.)
Characteristic equation has complex roots
  If the characteristic equation has complex roots, the general solution of the equation is $(A\cos(βt) + B\sin(βt))e^{αt}$, where $α = −a/2$, $β = \sqrt{b − a2/4}$. This function converges to 0 for all values of A and B, so that the equilibrium is globally stable, if and only if $α < 0$, or equivalently $a > 0$. (Remember that $\cos θ$ and $\sin θ$ lie between $+1$ and $−1$ for all values of $θ$.)

A: The system is stable, here's a graph of the associated vector field ($u = y$, $v = y'$)

