Fundamental Sets of Solutions — Question 9

Question 9 – Solution

Strategy. Changing the reference time changes the coordinates in a normalized fundamental pair, while the represented solution remains the same.

Step 1: Normalize at the reference time. Set $$\phi_s(t)=\cos(t-s),\qquad\psi_s(t)=\sin(t-s).$$ Both solve the equation and have the required data. Their data determinant is $1$, so any solution is $y(t)=a\cos(t-s)+b\sin(t-s)$, where $a=y(s)$ and $b=y'(s)$.

Step 2: Move the state. Evaluation and differentiation give $$\boxed{y(s+h)=a\cos h+b\sin h,\qquad y'(s+h)=-a\sin h+b\cos h.}$$

Step 3: Compose and reverse. Applying the same formulas with increment $k$ yields first component $$a(\cos h\cos k-\sin h\sin k)+b(\sin h\cos k+\cos h\sin k).$$ The addition formulas make this $a\cos(h+k)+b\sin(h+k)$; the second component is $-a\sin(h+k)+b\cos(h+k)$. Taking $k=-h$ gives $(a,b)$. Squaring the two components and adding cancels the cross terms and gives $a^2+b^2$.

Step 4: Locate the first turning point. Here $y=3\cos t+4\sin t$ and $y'=-3\sin t+4\cos t$. The first positive zero of $y'$ is $$\boxed{t_* =\arctan(4/3),\qquad y(t_*)=5.}$$ Indeed $\cos t_*=3/5$, $\sin t_*=4/5$, and the slope is positive before this time. In coordinates $(y,y')$, the state travels clockwise on the circle of radius $5$: at $(5,0)$ its derivative is $(0,-5)$.