Final Thoughts — Question 10

Question 10 – Solution

Strategy. Transform both the dependent variable and time. The chain-rule factor is essential when comparing numerical rates in different units.

Step 1: Transform the IVP. Since $u(\tau)=P(5\tau)/50$, $$\frac{du}{d\tau}=\frac 5{50}P'(5\tau) =\frac 5{50}\cdot\frac 15(50u)(1-u)=u(1-u).$$ The initial value is $u(0)=10/50=1/5$. Hence $$\boxed{u'=u(1-u),\qquad u(0)=1/5.}$$

Step 2: Verify and return to original units. The supplied function satisfies $$u'=\frac{4e^{-\tau}}{(1+4e^{-\tau})^2}=u(1-u),\qquad u(0)=1/5.$$ Substituting $\tau=t/5$ gives $$\boxed{P(t)=\frac{50}{1+4e^{-t/5}}.}$$ As a direct check, $P'=40e^{-t/5}/(1+4e^{-t/5})^2$ agrees with the original right-hand side and $P(0)=10$.

Step 3: First passage to half capacity. The candidate increases strictly and reaches $P=25$ when $u=1/2$. Therefore $$1+4e^{-\tau}=2\quad\Longrightarrow\quad \boxed{\tau=\ln 4,\qquad t=5\ln 4\text{ days}.}$$

Step 4: Same prediction, different labels. One unit of $\tau$ is five days, and one unit of $u$ is fifty population units. Omitting the factor $5$ in $du/d\tau=(5/50)P'(5\tau)$ would incorrectly give $u'=u(1-u)/5$ and misrepresent the normalized time scale. A change of units must transform rates as well as values.