Complex Roots — Question 3

Question 3 – Solution

Strategy. Differentiate the entire product, including the exponential factor, before locating peaks.

Step 1: Verify the complex-root model. The roots $-1\pm 2i$ give $\boxed{y''+2y'+5y=0}$. Differentiation yields $y'=e^{-t}(-\cos 2t-2\sin 2t)$, so $y(0)=1$ and $y'(0)=-1$. Substituting a second derivative verifies the equation.

Step 2: Locate and classify the extrema. Let $\theta=\arctan(1/2)\in(0,\pi/2)$. The stationary condition is $\cos 2t+2\sin 2t=0$, giving $$\boxed{t_k=(k\pi-\theta)/2,\qquad k=1,2,\ldots}.$$ At these times $\cos 2t_k=(-1)^k\cos\theta=(-1)^k2/\sqrt 5$. Since $y'=0$, the ODE gives $y''=-5y$. Thus odd $k$ give negative minima and even $k$ give positive maxima. The first minimum is at $(\pi-\theta)/2$; the first positive-time maximum is at $\pi-\theta/2$.

Step 3: Compare extrema with envelope contacts. Equality $|y|=e^{-t}$ requires $|\cos 2t|=1$, or $t=n\pi/2$ for $n\ge 0$. At those times $y'=-y\ne 0$, so they are not stationary points. The response does share the tangent of the corresponding envelope there.

At every stationary point, $$\boxed{|y(t_k)|=\frac 2{\sqrt 5}e^{-t_k}<e^{-t_k}}.$$ The decreasing envelope shifts the extrema away from cosine peaks.

Step 4: Compare like-signed peaks. Consecutive positive local maxima have indices $k$ and $k+2$, so their time separation is $\pi$. Their common trigonometric factor cancels, giving the height ratio $\boxed{e^{-\pi}}$. This is the envelope decay over a full cycle, even though the maxima occur strictly below the envelopes. At $t=0$ the solution has a one-sided endpoint maximum; it is not a positive-time stationary maximum.