Euler Method Dd2F2B

1. **Problem 1: Use Euler's method to approximate $y(0.2)$ and $y(0.4)$ for the initial value problem** Given: $$y' = 1 + 2x - 2y, \quad y(0) = -4$$ Step size $h = 0.2$ 2. **Euler's method formula:** $$y_{n+1} = y_n + h f(x_n,y_n)$$ where $f(x,y) = y' = 1 + 2x - 2y$ 3. **Calculate $y(0.2)$:** - Initial values: $x_0 = 0$, $y_0 = -4$ - Compute $f(x_0,y_0) = 1 + 2(0) - 2(-4) = 1 + 0 + 8 = 9$ - Apply Euler's method: $$y_1 = y_0 + h f(x_0,y_0) = -4 + 0.2 \times 9 = -4 + 1.8 = -2.2$$ 4. **Calculate $y(0.4)$:** - Next step: $x_1 = 0.2$, $y_1 = -2.2$ - Compute $f(x_1,y_1) = 1 + 2(0.2) - 2(-2.2) = 1 + 0.4 + 4.4 = 5.8$ - Apply Euler's method: $$y_2 = y_1 + h f(x_1,y_1) = -2.2 + 0.2 \times 5.8 = -2.2 + 1.16 = -1.04$$ 5. **Final approximations:** $$y(0.2) \approx -2.2, \quad y(0.4) \approx -1.04$$ --- 6. **Problem 2: Solve the differential equation** $$\frac{du}{dt} = e^{3.6t - 1.3u}, \quad u(0) = 3.3$$ 7. **Separate variables:** $$\frac{du}{dt} = e^{3.6t} e^{-1.3u} \implies e^{1.3u} du = e^{3.6t} dt$$ 8. **Rewrite as:** $$e^{1.3u} du = e^{3.6t} dt$$ 9. **Integrate both sides:** $$\int e^{1.3u} du = \int e^{3.6t} dt$$ 10. **Integrate:** $$\frac{1}{1.3} e^{1.3u} = \frac{1}{3.6} e^{3.6t} + C$$ 11. **Multiply both sides by $1.3$:** $$e^{1.3u} = \frac{1.3}{3.6} e^{3.6t} + 1.3 C$$ 12. **Apply initial condition $u(0) = 3.3$:** $$e^{1.3 \times 3.3} = \frac{1.3}{3.6} e^{0} + 1.3 C \implies e^{4.29} = \frac{1.3}{3.6} + 1.3 C$$ 13. **Solve for $C$:** $$1.3 C = e^{4.29} - \frac{1.3}{3.6} \implies C = \frac{e^{4.29} - \frac{1.3}{3.6}}{1.3}$$ 14. **Final implicit solution:** $$e^{1.3u} = \frac{1.3}{3.6} e^{3.6t} + e^{4.29} - \frac{1.3}{3.6}$$ 15. **Solve for $u$ explicitly:** $$u = \frac{1}{1.3} \ln \left( \frac{1.3}{3.6} e^{3.6t} + e^{4.29} - \frac{1.3}{3.6} \right)$$