Complex Roots — Question 6

Question 6 – Solution

Strategy. Remove the exponential factor and scale the remaining derivative so the two trigonometric coordinates have equal amplitudes.

Step 1: Derive the transformed equations. Directly, $u'=e^{at}(y'+ay)=bv$. Also, $$v'=\frac{e^{at}}b(y''+2ay'+a^2y) =\frac{e^{at}}b(-b^2y)=-bu.$$ The original equation is essential in the second equality.

Step 2: Identify the invariant and amplitude. Differentiating gives $(u^2+v^2)'=2u(bv)+2v(-bu)=0$. At zero, $$\boxed{u^2+v^2=Y^2+\frac{(V+aY)^2}{b^2}=R^2}.$$ The real solution has trigonometric coefficients $A=Y$, $B=(V+aY)/b$, so $R=\sqrt{A^2+B^2}$ is exactly its trigonometric amplitude.

Step 3: Trace the concrete circle. For the stated data, $A=1$, $B=0$. Thus $$\boxed{y=e^{-t}\cos 2t},\qquad u=\cos 2t,\quad v=-\sin 2t.$$ The initial point is $(1,0)$ and its velocity is $(0,-2)$, so motion is clockwise. At times $0,\pi/4,\pi/2,3\pi/4$ the points are $(1,0),(0,-1),(-1,0),(0,1)$, respectively; the cycle closes at $t=\pi$.

Step 4: Translate back to the original quantity. The transformation removed $e^{-at}$: the original response satisfies $|y|\le Re^{-at}$. Its amplitude decays for $a>0$, is constant for $a=0$, and grows for $a<0$. A circle in $(u,v)$ describes a normalized oscillation, not constant original displacement amplitude or a claim about physical energy.

Because $b>0$, the sum of squares $R^2$ is zero exactly when $Y=0$ and $V+aY=0$, hence exactly when $Y=V=0$. These data give $y\equiv 0$, consistent with initial-value uniqueness.