1. **Problem (九): Solve the differential equation** $s'(t) = 0.07s(t) + 40000$ with initial condition $s(0) = 1000000$. 2. **Step 1: Identify the type of differential equation.** This is a first-order linear ordinary differential equation of the form: $$ s'(t) - 0.07s(t) = 40000 $$ 3. **Step 2: Find the integrating factor (IF).** The integrating factor is: $$ \mu(t) = e^{-0.07t} $$ 4. **Step 3: Multiply both sides by the integrating factor:** $$ e^{-0.07t}s'(t) - 0.07e^{-0.07t}s(t) = 40000e^{-0.07t} $$ This simplifies to: $$ \frac{d}{dt}\left(s(t)e^{-0.07t}\right) = 40000e^{-0.07t} $$ 5. **Step 4: Integrate both sides with respect to $t$:** $$ s(t)e^{-0.07t} = \int 40000e^{-0.07t} dt + C $$ 6. **Step 5: Compute the integral:** $$ \int 40000e^{-0.07t} dt = 40000 \times \left(-\frac{1}{0.07}\right)e^{-0.07t} + C = -\frac{40000}{0.07}e^{-0.07t} + C $$ 7. **Step 6: Substitute back:** $$ s(t)e^{-0.07t} = -\frac{40000}{0.07}e^{-0.07t} + C $$ 8. **Step 7: Multiply both sides by $e^{0.07t}$:** $$ s(t) = -\frac{40000}{0.07} + Ce^{0.07t} $$ 9. **Step 8: Use initial condition $s(0) = 1000000$ to find $C$:** $$ 1000000 = -\frac{40000}{0.07} + C \Rightarrow C = 1000000 + \frac{40000}{0.07} $$ Calculate $\frac{40000}{0.07} = 571428.5714$ approximately. So, $$ C \approx 1000000 + 571428.5714 = 1571428.5714 $$ 10. **Final solution:** $$ s(t) = -571428.5714 + 1571428.5714 e^{0.07t} $$ --- 11. **Problem (十): Use 1st and 2nd degree Taylor expansions to approximate $\sqrt[3]{27.03}$** 12. **Step 1: Define the function and expansion point.** Let $$ f(x) = x^{1/3} $$ We approximate near $a=27$ because $27^{1/3} = 3$. 13. **Step 2: Compute derivatives at $a=27$:** $$ f'(x) = \frac{1}{3}x^{-2/3} $$ $$ f''(x) = -\frac{2}{9}x^{-5/3} $$ Evaluate at $x=27$: $$ f'(27) = \frac{1}{3} \times 27^{-2/3} = \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} $$ $$ f''(27) = -\frac{2}{9} \times 27^{-5/3} = -\frac{2}{9} \times \frac{1}{243} = -\frac{2}{2187} $$ 14. **Step 3: Calculate $h = 27.03 - 27 = 0.03$** 15. **Step 4: 1st degree Taylor approximation:** $$ f(27.03) \approx f(27) + f'(27)h = 3 + \frac{1}{27} \times 0.03 = 3 + 0.0011111 = 3.0011111 $$ 16. **Step 5: 2nd degree Taylor approximation:** $$ f(27.03) \approx f(27) + f'(27)h + \frac{f''(27)}{2}h^2 = 3 + \frac{1}{27} \times 0.03 + \frac{-2/2187}{2} \times (0.03)^2 $$ Calculate: $$ \frac{-2}{2187} \times \frac{1}{2} = -\frac{1}{2187} $$ $$ (0.03)^2 = 0.0009 $$ So, $$ 3 + 0.0011111 - \frac{1}{2187} \times 0.0009 = 3.0011111 - 0.000000411 = 3.0011107 $$ 17. **Final approximations:** - 1st degree: $3.0011$ - 2nd degree: $3.0011$ (more precise)