1. We are given the initial-value problem $$y' = x^2 y - \frac{1}{2} y^2,\quad y(0)=9$$ and we need to estimate $$y(1)$$ using Euler's method with step size $$h=0.2$$. 2. Recall Euler's method formula: $$y_{n+1} = y_n + h f(x_n,y_n)$$ where $$f(x,y) = x^2 y - \frac{1}{2} y^2$$. 3. Start with $$x_0=0$$ and $$y_0=9$$. 4. Compute each step: Step 1: $$x_1=0.2$$ $$f(x_0,y_0) = (0)^2 \times 9 - \frac{1}{2} \times 9^2 = 0 - 40.5 = -40.5$$ $$y_1 = 9 + 0.2 \times (-40.5) = 9 - 8.1 = 0.9$$ Step 2: $$x_2=0.4$$ $$f(x_1,y_1) = (0.2)^2 \times 0.9 - \frac{1}{2} \times (0.9)^2 = 0.04 \times 0.9 - 0.5 \times 0.81 = 0.036 - 0.405 = -0.369$$ $$y_2 = 0.9 + 0.2 \times (-0.369) = 0.9 - 0.0738 = 0.8262$$ Step 3: $$x_3=0.6$$ $$f(x_2,y_2) = (0.4)^2 \times 0.8262 - \frac{1}{2} \times (0.8262)^2 = 0.16 \times 0.8262 - 0.5 \times 0.6826 = 0.13219 - 0.3413 = -0.20911$$ $$y_3 = 0.8262 + 0.2 \times (-0.20911) = 0.8262 - 0.04182 = 0.78438$$ Step 4: $$x_4=0.8$$ $$f(x_3,y_3) = (0.6)^2 \times 0.78438 - \frac{1}{2} \times (0.78438)^2 = 0.36 \times 0.78438 - 0.5 \times 0.6153 = 0.2824 - 0.3077 = -0.0253$$ $$y_4 = 0.78438 + 0.2 \times (-0.0253) = 0.78438 - 0.00506 = 0.77932$$ Step 5: $$x_5=1.0$$ $$f(x_4,y_4) = (0.8)^2 \times 0.77932 - \frac{1}{2} \times (0.77932)^2 = 0.64 \times 0.77932 - 0.5 \times 0.6073 = 0.4980 - 0.3036 = 0.1944$$ $$y_5 = 0.77932 + 0.2 \times 0.1944 = 0.77932 + 0.03888 = 0.8182$$ 5. Rounded to four decimal places, the estimate is: $$\boxed{0.8182}$$