Complex Roots — Question 5

Question 5 – Solution

Strategy. Separate the exponential amplitude from the trigonometric zero pattern; growth of the former does not erase zeros of the latter.

Step 1: Find the universal complex-root form. The polynomial is $(r+a)^2+9$, so the roots are $\boxed{-a\pm 3i}$. Hence $$y=e^{-at}(A\cos 3t+B\sin 3t).$$ For a nontrivial solution, $R=\sqrt{A^2+B^2}>0$ and some phase $\delta$ give $y=Re^{-at}\cos(3t-\delta)$.

Step 2: Classify decay and boundedness. If $a>0$, $|y|\le Re^{-at}\to 0$. At $a=0$, all solutions are bounded, and every nontrivial one is periodic with least positive period $2\pi/3$ and does not tend to zero. If $a<0$, the magnitudes at trigonometric peaks grow without bound. Therefore $$\boxed{\text{all solutions decay}\iff a>0},\qquad \boxed{\text{all solutions are bounded}\iff a\ge 0}.$$ The zero solution is bounded and decays for every parameter value.

Step 3: Track the zeros and their slopes. Zeros occur when $3t-\delta=\pi/2+n\pi$, so $$t_n=\frac{\delta+\pi/2+n\pi}{3},$$ with all integers $n$ giving nonnegative times retained. There are infinitely many, spaced by $\boxed{\pi/3}$. At each zero, $$y'(t_n)=-3Re^{-at_n}\sin(\pi/2+n\pi)\ne 0.$$ Thus every zero is simple and the sign changes there. No nontrivial solution is eventually strictly positive or strictly negative, even when its amplitude decays.

Step 4: State growth without an incorrect limit. For $a<0$, choose large integers $n$ and set $s_n=(\delta+2n\pi)/3$. Then $y(s_n)=Re^{-as_n}\to+\infty$. At $u_n=(\delta+(2n+1)\pi)/3$, $y(u_n)=-Re^{-au_n}\to-\infty$. But along the zero sequence, $|y(t_n)|=0$.

Consequently the solution has unbounded magnitude, $\limsup y=+\infty$ and $\liminf y=-\infty$, while $\liminf |y|=0$. The limit $|y(t)|\to\infty$ is false. A growing envelope describes arbitrarily large excursions, not a lower bound on the response at every late time.