How to solve $y''+by'+y=0$,$ b\geq 0$; for what b does $y(t)$ have $\lim\limits_{t\to\infty} y(t)$? How to solve $$y''+by'+y=0$$ for $b\geq0$ (just need a quick reminder)? For what $b$ does $\lim\limits_{t \to \infty} y(t)$ exist (this is the more important question to me)?
 A: If you have a equation like a $y''+ b y'+ c = 0$ , you can find the solution using the associated quadratic polynomial to the differential equation : $D^2 + b D + C = 0$. So, resolve it with the quadratic formula
$$
D = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$
You could have complex roots so, in your solutions should appear $\cos x$, $\sin x$ and then, you can propose the solutions of your differential equation, and the form could  be:
$$
y(t)= k e^{D t}
$$
so, analyze $\lim\limits_{t\to \infty}y(t)$. Then look how behaves exponential function.
Regards, 
A: There is a mild exception that we discuss later, but in general the solutions of the differential equation have the shape $Ae^{\lambda_1 t}+Be^{\lambda_2 t}$ where $\lambda_1$ and $\lambda_2$ are the (possibly non-real) roots of the equation $\lambda^2+b\lambda+1=0$. 
By the Quadratic Formula, the roots of this quadratic equation are $\frac{-b \pm\sqrt{b^2-4}}{2}$.   
If $b=0$ we are in trouble, the solutions of the DE are linear combinations of $\sin t$ and $\cos t$, and the limit as $t \to \infty$ does not exist, except in the case of the trivial solution $y=0$. 
If $0&ltb&lt2$, the solutions of the DE are of the shape $e^{-bt/2}$ times a linear combination of sines and cosines, and the $e^{-bt/2}$ term forces the limits to be $0$. And if $b>2$, the two roots $\lambda_1$ and $\lambda_2$ are negative, so again all solutions $\to 0$ as $t\to\infty$.
In the case $b=2$, the equation $\lambda^2+b\lambda+1=0$ has the double root $\lambda=-1$, and the solutions of the differential equation are linear combinations of $e^{-t}$ and $te^{-t}$. Both of these approach $0$ as $t\to \infty$.
You did not ask about $b&lt0$. In this case, all solutions except the trivial solution $y=0$ blow up as $t\to\infty$.
