1. **State the problem:** We want to approximate the solution to the differential equation $$y' = x + y$$ with initial condition $$y(0) = 0$$ using Euler's method with step size $$h = 0.2$$ for 6 steps. 2. **Recall Euler's method formula:** $$y_{n+1} = y_n + h f(x_n, y_n)$$ where $$f(x,y) = x + y$$ is the derivative function. 3. **Initialize:** $$x_0 = 0, y_0 = 0$$ 4. **Calculate each step:** - Step 1: $$x_1 = x_0 + h = 0 + 0.2 = 0.2$$ $$y_1 = y_0 + h (x_0 + y_0) = 0 + 0.2 (0 + 0) = 0$$ - Step 2: $$x_2 = 0.4$$ $$y_2 = y_1 + 0.2 (x_1 + y_1) = 0 + 0.2 (0.2 + 0) = 0.04$$ - Step 3: $$x_3 = 0.6$$ $$y_3 = y_2 + 0.2 (x_2 + y_2) = 0.04 + 0.2 (0.4 + 0.04) = 0.04 + 0.2 (0.44) = 0.04 + 0.088 = 0.128$$ - Step 4: $$x_4 = 0.8$$ $$y_4 = y_3 + 0.2 (x_3 + y_3) = 0.128 + 0.2 (0.6 + 0.128) = 0.128 + 0.2 (0.728) = 0.128 + 0.1456 = 0.2736$$ - Step 5: $$x_5 = 1.0$$ $$y_5 = y_4 + 0.2 (x_4 + y_4) = 0.2736 + 0.2 (0.8 + 0.2736) = 0.2736 + 0.2 (1.0736) = 0.2736 + 0.21472 = 0.48832$$ - Step 6: $$x_6 = 1.2$$ $$y_6 = y_5 + 0.2 (x_5 + y_5) = 0.48832 + 0.2 (1.0 + 0.48832) = 0.48832 + 0.2 (1.48832) = 0.48832 + 0.297664 = 0.785984$$ 5. **Final approximate values:** $$y(0.2) \approx 0$$ $$y(0.4) \approx 0.04$$ $$y(0.6) \approx 0.128$$ $$y(0.8) \approx 0.2736$$ $$y(1.0) \approx 0.48832$$ $$y(1.2) \approx 0.785984$$ This completes 6 steps of Euler's method with step size 0.2.