Complex Roots — Question 4

Question 4 – Solution

Strategy. Use the interval between like-signed extrema for frequency and their height ratio for the real part of the roots.

Step 1: Extract frequency and decay. Write $y=Re^{\alpha t}\cos(\beta t-\delta)$. Stationarity requires $\alpha\cos(\beta t-\delta)-\beta\sin(\beta t-\delta)=0$. Successive extrema have phase separation $\pi$ and alternate sign, so consecutive positive maxima are separated by $2\pi/\beta$.

The observed separation is 2, giving $\boxed{\beta=\pi}$. The trigonometric factors at those maxima agree, so $1/4=e^{2\alpha}$ and $$\boxed{\alpha=-\ln 2}.$$

Step 2: Recover the real coefficients. Let $k=\ln 2>0$. The characteristic polynomial is $(r+k)^2+\pi^2$, hence $$\boxed{y''+2k y'+(k^2+\pi^2)y=0}.$$

Step 3: Reconstruct from value and zero slope. Using $x=t-1$, write $y=e^{-kx}(A\cos\pi x+B\sin\pi x)$. The peak gives $A=4$ and $-kA+\pi B=0$, so $$\boxed{y=e^{-k(t-1)}[4\cos(\pi(t-1))+\frac{4k}{\pi}\sin(\pi(t-1))]}.$$ At $t=3$, $x=2$, and $y=4e^{-2k}=1$, $y'=0$. At either point the equation gives $y''=-(k^2+\pi^2)y<0$, verifying strict positive local maxima. The solution is unique because its value and slope at 1 are specified for a regular linear equation.

Step 4: Identify what consecutiveness contributes. Without that condition, the time interval 2 could contain $m$ full cycles, where $m$ is any positive integer. Then $\boxed{\beta=m\pi}$ while the height ratio still fixes $\alpha=-\ln 2$. Each choice gives a different equation and a solution with the stated peaks, but for $m>1$ there are additional positive maxima between them. The observed heights alone do not resolve the missing cycle count.