1. The problem involves two differential equations describing population growth and decay: (E₁): $y' = 0.022y(20 - y)$ and (E₂): $y' = -0.44y + 0.022$ Given the solution for (E₁) as $f(t) = \frac{20}{1 + 1999e^{-0.44t}}$, and $u = \frac{1}{f}$. 2. For (E₁), the logistic growth model is used: $y' = ky(M - y)$ where $k=0.022$ and carrying capacity $M=20$. 3. The solution $f(t)$ is a logistic function with initial condition embedded in the constant 1999. 4. For (E₂), rewrite as $y' + 0.44y = 0.022$, a linear first-order ODE. 5. Use integrating factor $\mu(t) = e^{0.44t}$: $$\frac{d}{dt}(ye^{0.44t}) = 0.022e^{0.44t}$$ 6. Integrate both sides: $$ye^{0.44t} = \int 0.022e^{0.44t} dt = 0.022 \times \frac{e^{0.44t}}{0.44} + C = 0.05e^{0.44t} + C$$ 7. Solve for $y$: $$y = 0.05 + Ce^{-0.44t}$$ 8. Using initial condition $y(0) = y_0$, find $C = y_0 - 0.05$. 9. Thus, the general solution for (E₂) is: $$y(t) = 0.05 + (y_0 - 0.05)e^{-0.44t}$$ 10. The function $f(t)$ given matches the logistic solution for (E₁). Final answers: - For (E₁): $f(t) = \frac{20}{1 + 1999e^{-0.44t}}$ - For (E₂): $y(t) = 0.05 + (y_0 - 0.05)e^{-0.44t}$