1. **State the problem:** We want to solve the differential equation $$y' = 1 - y$$ with initial condition $$y(0) = 0$$ using the modified Euler's method (also known as Heun's method) and find the values of $$y$$ at $$x = 0.1, 0.2, 0.3$$. 2. **Recall the modified Euler's method formula:** Given $$y' = f(x,y)$$ and step size $$h$$, the method updates $$y$$ as: $$ \begin{aligned} k_1 &= f(x_n, y_n) \\ k_2 &= f(x_n + h, y_n + h k_1) \\ y_{n+1} &= y_n + \frac{h}{2}(k_1 + k_2) \end{aligned} $$ This method improves accuracy by averaging slopes at the beginning and end of the interval. 3. **Set up initial values:** - Initial condition: $$x_0 = 0, y_0 = 0$$ - Step size: $$h = 0.1$$ - Function: $$f(x,y) = 1 - y$$ 4. **Calculate $$y_1$$ at $$x_1 = 0.1$$:** - $$k_1 = f(0,0) = 1 - 0 = 1$$ - $$k_2 = f(0.1, 0 + 0.1 \times 1) = f(0.1, 0.1) = 1 - 0.1 = 0.9$$ - $$y_1 = 0 + \frac{0.1}{2}(1 + 0.9) = 0.05 \times 1.9 = 0.095$$ 5. **Calculate $$y_2$$ at $$x_2 = 0.2$$:** - $$k_1 = f(0.1, 0.095) = 1 - 0.095 = 0.905$$ - $$k_2 = f(0.2, 0.095 + 0.1 \times 0.905) = f(0.2, 0.1855) = 1 - 0.1855 = 0.8145$$ - $$y_2 = 0.095 + \frac{0.1}{2}(0.905 + 0.8145) = 0.095 + 0.05 \times 1.7195 = 0.095 + 0.085975 = 0.180975$$ 6. **Calculate $$y_3$$ at $$x_3 = 0.3$$:** - $$k_1 = f(0.2, 0.180975) = 1 - 0.180975 = 0.819025$$ - $$k_2 = f(0.3, 0.180975 + 0.1 \times 0.819025) = f(0.3, 0.2628775) = 1 - 0.2628775 = 0.7371225$$ - $$y_3 = 0.180975 + \frac{0.1}{2}(0.819025 + 0.7371225) = 0.180975 + 0.05 \times 1.5561475 = 0.180975 + 0.077807375 = 0.258782375$$ 7. **Final answers:** - $$y(0.1) \approx 0.095$$ - $$y(0.2) \approx 0.181$$ - $$y(0.3) \approx 0.259$$ These values approximate the solution to the differential equation at the specified points using the modified Euler's method.