Repeated Roots — Question 10

Question 10 – Solution

Strategy. Scale time by the repeated decay rate, then compare the performance requirement with the independent constraint on the initial acceleration.

Step 1: Solve and establish monotonicity. The repeated root is $-\lambda$. The data give $$\boxed{y=(1+\lambda t)e^{-\lambda t}},\qquad y'=-\lambda^2t e^{-\lambda t}.$$ The response is positive, strictly decreasing for $t>0$, starts at 1 and tends to zero. Therefore crossing the tolerance once is enough to guarantee it thereafter.

Step 2: Reduce the crossing calculation to one dimensionless number. Let $f(z)=(1+z)e^{-z}$ for $z\ge 0$. Then $f(0)=1$, $f(z)\to 0$ and $f'(z)=-ze^{-z}<0$ for $z>0$. The intermediate value theorem and strict decrease give exactly one positive root $f(z_*)=0.05$.

Direct evaluation gives $f(4.7438)>0.05$ and $f(4.7439)<0.05$. Thus $$\boxed{4.7438<z_*<4.7439},\qquad t_{\mathrm{settle}}=z_*/\lambda.$$ Numerically $z_*\approx 4.743865$; the bracket, rather than a rounded equality, supports the feasibility comparisons below.

Step 3: Test simultaneous feasibility. Settling by time 2 requires $z_*/\lambda\le 2$, or $\lambda\ge z_*/2>2.3719$. But the equation at zero gives $y''(0)=-\lambda^2$, so the acceleration constraint requires $\lambda\le 2$. These ranges do not intersect: $\boxed{\text{the two requirements are incompatible}}$.

At the largest admissible value $\lambda=2$, the response at the deadline is $y(2)=5e^{-4}\approx 0.091578>0.05$, confirming the failure directly.

Step 4: Optimize within the allowed family. The settling time $z_*/\lambda$ strictly decreases as $\lambda$ increases. Under $0<\lambda\le 2$, it is minimized at $\lambda=2$, giving $$\boxed{t_{\min}=z_*/2\approx 2.371932}.$$ This is later than the required time 2. Larger $\lambda$ would improve the settling time but violate the initial-acceleration bound. The conclusion optimizes this explicitly specified family and constraint, not every possible differential-equation model or control law.