Equilibrium Solutions — Question 2

Question 2 – Solution

Strategy. Preserve the sign across a squared factor; use one-sided attraction and repulsion when the derivative test is inconclusive.

Step 1: Build the sign chart. The equilibria are $-1$ and $2$. Away from them, $(y-2)^2>0$, so the sign of $f$ is the opposite of the sign of $y+1$: $$\begin{array}{c|ccc} y&(-\infty,-1)&(-1,2)&(2,\infty)\\\hline f(y)&+&-&- \end{array}$$

Step 2: Classify with the sides stated. The equilibrium $\boxed{-1\text{ is asymptotically stable}}$. At $2$, solutions above it decrease toward it, while those just below decrease away toward $-1$. Hence $$\boxed{2\text{ is semistable: attracting from above, repelling from below}.}$$ It is unstable in the two-sided stability sense: however close an initial value is below $2$, its trajectory eventually leaves, for example, the interval $(1,3)$.

Step 3: Track the three initial values. Uniqueness prevents crossing either equilibrium. The solution starting at $-2$ increases toward $-1$; that starting at $1$ decreases toward $-1$; that starting at $3$ decreases toward $2$. Each remains in a bounded closed interval, so the smooth equation has a global forward solution in each case. Monotone convergence and the nonzero-speed contradiction identify the stated limits.

Step 4: Interpret the derivative test. Differentiating gives $f'(y)=-(y-2)^2-2(y+1)(y-2)$, so $$f'(-1)=-9<0,\qquad f'(2)=0.$$ The negative derivative confirms attraction at $-1$. At $2$, the linearized equation contains no nonzero first-order term and makes no decision. The sign chart supplies the missing information. A squared factor is positive on both sides of its zero, so it touches zero without a sign reversal.