1. **State the problem:** Solve the differential equation $$\frac{dB}{dt} + 0.3B = 0.02B^2$$ with initial condition $$B(0) = 1$$. 2. **Identify the type of equation:** This is a Bernoulli differential equation of the form $$\frac{dy}{dt} + P(t)y = Q(t)y^n$$ where here $$y = B$$, $$P(t) = 0.3$$, $$Q(t) = 0.02$$, and $$n=2$$. 3. **Transform the equation:** Divide both sides by $$B^2$$ to get $$\frac{1}{B^2}\frac{dB}{dt} + 0.3\frac{1}{B} = 0.02$$ Rewrite $$\frac{dB}{dt}$$ as $$\frac{dB}{dt} = B^2 \frac{d}{dt}\left(\frac{1}{B}\right)$$ using the substitution $$v = \frac{1}{B}$$. 4. **Substitute:** Let $$v = \frac{1}{B}$$, then $$\frac{dv}{dt} = -\frac{1}{B^2}\frac{dB}{dt}$$. Multiply original equation by $$\frac{1}{B^2}$$: $$\frac{1}{B^2}\frac{dB}{dt} + 0.3\frac{1}{B} = 0.02$$ Substitute $$\frac{1}{B^2}\frac{dB}{dt} = -\frac{dv}{dt}$$ and $$\frac{1}{B} = v$$: $$-\frac{dv}{dt} + 0.3v = 0.02$$ 5. **Rearrange:** $$\frac{dv}{dt} - 0.3v = -0.02$$ This is a linear first-order ODE in $$v$$. 6. **Solve the linear ODE:** The integrating factor is $$\mu(t) = e^{-0.3t}$$ Multiply both sides: $$e^{-0.3t}\frac{dv}{dt} - 0.3e^{-0.3t}v = -0.02e^{-0.3t}$$ Left side is derivative: $$\frac{d}{dt}\left(v e^{-0.3t}\right) = -0.02 e^{-0.3t}$$ 7. **Integrate both sides:** $$\int \frac{d}{dt}\left(v e^{-0.3t}\right) dt = \int -0.02 e^{-0.3t} dt$$ $$v e^{-0.3t} = -0.02 \int e^{-0.3t} dt + C$$ Integral: $$\int e^{-0.3t} dt = \frac{e^{-0.3t}}{-0.3} = -\frac{1}{0.3} e^{-0.3t}$$ So: $$v e^{-0.3t} = -0.02 \left(-\frac{1}{0.3} e^{-0.3t}\right) + C = \frac{0.02}{0.3} e^{-0.3t} + C$$ 8. **Multiply both sides by $$e^{0.3t}$$:** $$v = \frac{0.02}{0.3} + C e^{0.3t}$$ 9. **Apply initial condition:** At $$t=0$$, $$B(0) = 1$$ so $$v(0) = \frac{1}{B(0)} = 1$$. Substitute: $$1 = \frac{0.02}{0.3} + C e^{0} = \frac{0.02}{0.3} + C$$ Calculate $$\frac{0.02}{0.3} = \frac{2}{30} = \frac{1}{15} \approx 0.0667$$ So: $$C = 1 - 0.0667 = 0.9333$$ 10. **Final solution for $$v$$:** $$v = \frac{1}{15} + 0.9333 e^{0.3t}$$ Recall $$v = \frac{1}{B}$$, so $$B(t) = \frac{1}{\frac{1}{15} + 0.9333 e^{0.3t}}$$ This function models the biomass concentration over time. --- **Final answer:** $$\boxed{B(t) = \frac{1}{\frac{1}{15} + 0.9333 e^{0.3t}}}$$