Differential Equations: Which values of b = damping coefficient does the typical solution approach the equilibrium position most rapidly

Question

Here's where I'm at with it: $\lambda ^2+b\lambda +3=0$

$\lambda _1=\frac{\left(-b+\sqrt{b^2-12}\right)}{2},\:\lambda \:_2=\frac{\left(-b-\sqrt{b^2-12}\right)}{2}$

I do not know how to complete this problem. Help would be greatly appreciated.TIA

Answer 1

$b>0$ for it to be a damping term. If $b^2<12$ then the square root gives an imaginary term and hence it will oscillate.

$\lambda_2<\lambda_1$ so the second solution will die out quicker than the first. So it amounts to finding the $b$ that gives the largest negative value of $\lambda_1$.

$-\lambda_1$ falls monotonically from $b=\sqrt{12}$ to $\lambda_1=0$ at $b=\infty$ so $b=\sqrt{12}$ is the answer.