Equilibrium Solutions — Question 8

Question 8 – Solution

Strategy. A denominator zero splits the domain but is not an equilibrium. Test whether the selected trajectory reaches that boundary in finite time.

Step 1: Separate zeros from exclusions. The only equilibrium is $\boxed{y=1}$. The value $y=-2$ is outside the equation’s domain. The signs of $f$ are positive on $(-\infty,-2)$, negative on $(-2,1)$, and positive on $(1,\infty)$.

Thus $1$ is unstable: arrows point away on both sides. Also $f'(y)=3/(y+2)^2$, so $f'(1)=1/3>0$ confirms the classification.

Step 2: Integrate the selected branch. Starting at $0$, the solution decreases within $(-2,1)$. Separation gives $$dt=\frac{y+2}{y-1}\,dy,\qquad \boxed{t=y+3\ln(1-y)}.$$ The constant is zero from $y(0)=0$. On $-2<y\le 0$, the derivative of the right side with respect to $y$ is $(y+2)/(y-1)<0$. It is therefore invertible along this branch, and implicit differentiation verifies the original equation.

Step 3: Determine the endpoint. As $y\downarrow-2$, the time tends to $$\boxed{T=-2+3\ln 3\approx 1.296},\qquad \boxed{0\le t<T}.$$ This is the maximal forward range: the inverse exists until $T$, but at $T$ its limiting value is excluded. Within the branch, $$y(t)\to-2^+,\qquad y'(t)=\frac{y-1}{y+2}\to-\infty.$$

Step 4: Interpret the domain boundary. An attracting equilibrium must itself define a constant solution of the given equation. Here $-2$ does not, and the selected solution reaches the boundary in finite time with unbounded slope rather than converging to a valid equilibrium as $t\to\infty$.

Multiplication gives $(y+2)y'=y-1$, equivalent to the original only when $y\ne-2$. At $y=-2$ it would demand $0=-3$ for any finite derivative. Thus even the multiplied relation supplies no differentiable continuation through this boundary.