Complex Roots — Question 2

Question 2 – Solution

Strategy. Determine the sine and cosine coefficients before choosing the phase quadrant; then use the phase to locate zeros.

Step 1: Solve the real IVP. The roots are $-1\pm 2i$. For $y=e^{-t}(A\cos 2t+B\sin 2t)$, the data give $A=-1$ and $-A+2B=3$. Thus $B=1$ and $$\boxed{y=e^{-t}(-\cos 2t+\sin 2t)}.$$

Step 2: Choose the correct quadrant. Expansion of $R\cos(2t-\delta)$ gives $A=R\cos\delta$, $B=R\sin\delta$. Therefore $$R=\sqrt{A^2+B^2}=\sqrt 2,\qquad \cos\delta=-1/\sqrt 2,\quad\sin\delta=1/\sqrt 2.$$ The prescribed convention selects $\boxed{\delta=3\pi/4}$, yielding $$\boxed{y=\sqrt 2e^{-t}\cos(2t-3\pi/4)}.$$ At zero the value is $-1$, while the derivative is $\sqrt 2[-\cos(-3\pi/4)-2\sin(-3\pi/4)]=3$, as required.

Step 3: Enumerate the nonnegative zeros. The exponential is never zero. The first zero occurs when $2t-3\pi/4=-\pi/2$, so all nonnegative zeros are $$\boxed{t_n=\frac\pi 8+\frac{n\pi}{2},\qquad n=0,1,2,\ldots}.$$ At $t_0$, the derivative is $2\sqrt 2e^{-\pi/8}>0$, so the solution crosses from negative to positive. Successive zeros are separated by $\pi/2$, half the trigonometric period $\pi$. The decaying solution itself is not periodic.

Step 4: Resolve the tangent ambiguity. The equation $\tan\delta=-1$ does not distinguish quadrants II and IV. With positive $R=\sqrt 2$, the phase $-\pi/4$ gives $A=1$, $B=-1$, which is the negative of the required solution. It has value 1 and slope $-3$ at zero. The signs of both sine and cosine, together with the amplitude convention, determine the phase; a tangent value alone does not.