1. **State the problem:** We want to approximate the solution to the initial value problem $$y' = \frac{3}{x}(y^2 + y), \quad y(6) = 3$$ at points $x = 6.2, 6.4, 6.6, 6.8$ using Euler's method with step size $h = 0.2$. 2. **Euler's method formulas:** Given step size $h$, the recursion formulas are: $$x_{n+1} = x_n + h = x_n + 0.2$$ $$y_{n+1} = y_n + h \cdot f(x_n, y_n) = y_n + 0.2 \cdot \frac{3}{x_n}(y_n^2 + y_n) = y_n + \frac{0.6}{x_n}(y_n^2 + y_n)$$ 3. **Initial values:** $$x_0 = 6, \quad y_0 = 3$$ 4. **Calculate $y_1$ at $x_1 = 6.2$:** $$y_1 = y_0 + \frac{0.6}{x_0}(y_0^2 + y_0) = 3 + \frac{0.6}{6}(3^2 + 3) = 3 + \frac{0.6}{6}(9 + 3) = 3 + \frac{0.6}{6} \times 12$$ $$= 3 + 0.1 \times 12 = 3 + 1.2 = 4.2$$ 5. **Calculate $y_2$ at $x_2 = 6.4$:** $$y_2 = y_1 + \frac{0.6}{x_1}(y_1^2 + y_1) = 4.2 + \frac{0.6}{6.2}(4.2^2 + 4.2)$$ Calculate inside the parentheses: $$4.2^2 = 17.64, \quad 17.64 + 4.2 = 21.84$$ Then: $$y_2 = 4.2 + \frac{0.6}{6.2} \times 21.84 = 4.2 + 0.096774 \times 21.84 \approx 4.2 + 2.113 = 6.313$$ 6. **Calculate $y_3$ at $x_3 = 6.6$:** $$y_3 = y_2 + \frac{0.6}{x_2}(y_2^2 + y_2) = 6.313 + \frac{0.6}{6.4}(6.313^2 + 6.313)$$ Calculate inside the parentheses: $$6.313^2 \approx 39.85, \quad 39.85 + 6.313 = 46.163$$ Then: $$y_3 = 6.313 + \frac{0.6}{6.4} \times 46.163 = 6.313 + 0.09375 \times 46.163 \approx 6.313 + 4.324 = 10.637$$ 7. **Calculate $y_4$ at $x_4 = 6.8$:** $$y_4 = y_3 + \frac{0.6}{x_3}(y_3^2 + y_3) = 10.637 + \frac{0.6}{6.6}(10.637^2 + 10.637)$$ Calculate inside the parentheses: $$10.637^2 \approx 113.12, \quad 113.12 + 10.637 = 123.757$$ Then: $$y_4 = 10.637 + \frac{0.6}{6.6} \times 123.757 = 10.637 + 0.090909 \times 123.757 \approx 10.637 + 11.250 = 21.887$$ 8. **Summary table:** | n | $x_n$ | Euler's Method Approximation $y_n$ | |---|-------|------------------------------------| | 1 | 6.2 | 4.2 | | 2 | 6.4 | 6.313 | | 3 | 6.6 | 10.637 | | 4 | 6.8 | 21.887 | **Final answer:** $$y(6.2) \approx 4.2, \quad y(6.4) \approx 6.313, \quad y(6.6) \approx 10.637, \quad y(6.8) \approx 21.887$$