Complex Roots — Question 9

Question 9 – Solution

Strategy. Separate cancellation between lobes from decay of each lobe, then sum the absolute areas as a geometric series.

Step 1: Identify the equation and signs. The roots are $-1\pm i$, so $\boxed{y''+2y'+2y=0}$. The derivative is $e^{-t}(\cos t-\sin t)$, giving $y(0)=0$, $y'(0)=1$. On $(n\pi,(n+1)\pi)$, the sign is $(-1)^n$: positive first, negative next, and alternating thereafter.

Step 2: Integrate the signed response. Integration by parts twice, or differentiation of the result, gives $$F(t)=-\frac 12e^{-t}(\sin t+\cos t),\qquad F'=e^{-t}\sin t.$$ Since $F(t)\to 0$ as $t\to\infty$ and $F(0)=-1/2$, $$\boxed{I=1/2}.$$

Step 3: Sum the magnitudes without cancellation. The first positive lobe has area $$K=F(\pi)-F(0)=\frac{1+e^{-\pi}}2.$$ The substitution $t=n\pi+s$ shows that the absolute area of the $n$th lobe is $e^{-n\pi}K$. Thus the convergent geometric sum gives $$\boxed{J=\frac{K}{1-e^{-\pi}}=\frac{1+e^{-\pi}}{2(1-e^{-\pi})}>\frac 12}.$$ The negative lobes reduce $I$ but contribute positively to $J$. Infinitely many oscillations do not prevent absolute convergence because the lobe magnitudes decrease geometrically.

Step 4: Control the unobserved tail. Since $|\sin t|\le 1$, $$\boxed{\int_T^\infty |y(t)|\,dt\le\int_T^\infty e^{-t}\,dt=e^{-T}}.$$ At a half-cycle boundary, substitution by $n\pi$ preserves $|\sin|$, so the exact absolute tail is $\boxed{e^{-n\pi}J}$. The signed area of the second plotted lobe is $-e^{-\pi}K$, while its contribution to $J$ is $+e^{-\pi}K$.