Real & Distinct Roots — Question 2

Question 2 – Solution

Strategy. Express the initial data in the two modes and isolate the coefficient of the growing exponential.

Step 1: Resolve the data into modes. The characteristic polynomial is $(r-3)(r+2)$, so $y=Ae^{3t}+Be^{-2t}$. From $A+B=1$ and $3A-2B=v$, $$\boxed{A=(v+2)/5,\quad B=(3-v)/5}.$$ The nonzero root separation makes this data system invertible.

Step 2: Remove the growing mode exactly. If $A\ne 0$, the term $Ae^{3t}$ dominates and $|y|\to\infty$. Thus forward boundedness holds exactly when $A=0$, or $\boxed{v=-2}$. The selected solution is $\boxed{y=e^{-2t}}$ and tends to zero.

Step 3: Track a small slope perturbation. For $v=-2+\varepsilon$, $$y_\varepsilon=\frac{\varepsilon}{5}e^{3t} +\frac{5-\varepsilon}{5}e^{-2t}.$$ Both terms are positive. They are equal precisely when $e^{5t}=(5-\varepsilon)/\varepsilon$, giving $$\boxed{t_*=\frac 15\ln\frac{5-\varepsilon}{\varepsilon}}.$$ This is positive for the stated range of $\varepsilon$. Eventually $y_\varepsilon\to+\infty$, although its initial slope is negative.

Step 4: Quantify the delayed disagreement. For $\varepsilon=0.01$, $t_*=(\ln 499)/5\approx\boxed{1.243}$. The exact difference is $$y_\varepsilon-e^{-2t}=\frac{\varepsilon}{5}(e^{3t}-e^{-2t}).$$ It vanishes initially and is small with $\varepsilon$ on any fixed finite interval, but grows without bound as $t\to\infty$ for every fixed positive $\varepsilon$. Forward boundedness here requires exact cancellation of a mode; it is not robust to a nonzero slope error. The time $t_*$ compares mode magnitudes and is not the time of the trajectory’s minimum.