Definitions — Question 4

Question 4 – Solution

Strategy. Check the original equation before transforming it. Division by an expression involving the unknown can discard legitimate solutions.

Step 1: Constant solutions. If $y(x)=k$, then $y'=0$, so $$0=k(1-k)\quad\Longleftrightarrow\quad \boxed{k=0\text{ or }k=1.}$$ Both constant functions solve the equation on $\mathbb R$; these are its equilibrium solutions.

Step 2: Nonconstant verification. The quotient rule gives $$p'(x)=\frac{e^x(1+e^x)-e^{2x}}{(1+e^x)^2} =\frac{e^x}{(1+e^x)^2}.$$ Also $$p(1-p)=\frac{e^x}{1+e^x}\frac{1}{1+e^x}=p'.$$ The denominator is positive for every real $x$. Because $e^x$ ranges over $(0,\infty)$, $$\boxed{\operatorname{domain}(p)=\mathbb R,\qquad\operatorname{range}(p)=(0,1).}$$

Step 3: Restricted equivalence. Division is valid on an interval only if $y(x)\ne 0$ and $y(x)\ne 1$ everywhere on that interval. Thus it is valid for $p$, but is undefined for both equilibrium solutions. The equations are equivalent under the nonvanishing restriction, not for every original solution.

Step 4: Initial data and scope. Direct substitution yields $$\boxed{y_0(0)=0,\qquad y_1(0)=1,\qquad p(0)=\tfrac 12.}$$ Each is therefore a verified solution of the corresponding IVP. Testing three candidates does not exclude every other function: existence of a matching candidate and uniqueness of an IVP are logically different claims.