Direction Fields — Question 2

Question 2 – Solution

Strategy. Read the signs between horizontal zero-slope lines, then use nonintersection and a bounded-monotone argument.

Step 1: Field structure. The slope depends only on $y$, so it is the same along every horizontal row. Constant solutions require $y(2-y)=0$, giving $\boxed{y=0\text{ and }y=2}$. The sign is negative for $y<0$, positive for $0<y<2$, and negative for $y>2$.

Step 2: Confinement and monotonicity. Starting at $1$, a continuous solution could leave $(0,2)$ only by meeting one of the equilibrium solutions. The stated nonintersection fact excludes this. Therefore $$\boxed{0<y(x)<2,\qquad y'(x)>0\quad(x\ge 0).}$$ Since $y(0)=1$, the function is bounded below by $1$ on this half-line.

Step 3: Limit without an explicit formula. An increasing function bounded above by $2$ has a limit $L$ with $1\le L\le 2$. If $L<2$, continuity of $g(y)=y(2-y)$ gives $g(L)>0$ and, eventually, $y'(x)\ge g(L)/2=:c>0$. Integrating from a sufficiently large $X$ yields $$y(x)\ge y(X)+c(x-X),$$ contradicting the upper bound $2$. Thus $\boxed{\displaystyle\lim_{x\to\infty}y(x)=2}$. The limit is approached from below, without touching the equilibrium at any finite $x$.