1. **State the problem:** We want to solve the differential equation $$y' = x^2 + y^2$$ with initial condition $$y(0) = 1$$ using the Taylor series method of order four on the interval $$0 \leq x \leq 0.4$$ with step size $$h=0.2$$. 2. **Recall the Taylor series method formula:** $$ y(x+h) = y(x) + hy'(x) + \frac{h^2}{2!}y''(x) + \frac{h^3}{3!}y'''(x) + \frac{h^4}{4!}y^{(4)}(x) + \cdots $$ We will use terms up to order four. 3. **Calculate derivatives:** Given $$y' = f(x,y) = x^2 + y^2$$. - First derivative: $$y' = x^2 + y^2$$ - Second derivative: $$y'' = \frac{d}{dx}y' = \frac{d}{dx}(x^2 + y^2) = 2x + 2yy'$$ - Third derivative: $$y''' = \frac{d}{dx}y'' = \frac{d}{dx}(2x + 2yy') = 2 + 2(y'y' + y y'') = 2 + 2(y'^2 + y y'')$$ - Fourth derivative: $$y^{(4)} = \frac{d}{dx}y''' = \frac{d}{dx}\left(2 + 2(y'^2 + y y'')\right) = 2(2 y' y'' + y' y'' + y y''') = 2(3 y' y'' + y y''')$$ 4. **Evaluate derivatives at initial point $$x_0=0$$, $$y_0=1$$:** - $$y'_0 = 0^2 + 1^2 = 1$$ - $$y''_0 = 2(0) + 2(1)(1) = 2$$ - $$y'''_0 = 2 + 2(1^2 + 1 \times 2) = 2 + 2(1 + 2) = 2 + 6 = 8$$ - $$y^{(4)}_0 = 2(3 \times 1 \times 2 + 1 \times 8) = 2(6 + 8) = 2 \times 14 = 28$$ 5. **Apply Taylor series formula to find $$y_1 = y(0.2)$$:** $$ y_1 = y_0 + h y'_0 + \frac{h^2}{2} y''_0 + \frac{h^3}{6} y'''_0 + \frac{h^4}{24} y^{(4)}_0 $$ Substitute values: $$ y_1 = 1 + 0.2 \times 1 + \frac{0.2^2}{2} \times 2 + \frac{0.2^3}{6} \times 8 + \frac{0.2^4}{24} \times 28 $$ Calculate each term: - $$0.2 \times 1 = 0.2$$ - $$\frac{0.04}{2} \times 2 = 0.04$$ - $$\frac{0.008}{6} \times 8 = \frac{0.008 \times 8}{6} = \frac{0.064}{6} \approx 0.0106667$$ - $$\frac{0.0016}{24} \times 28 = \frac{0.0016 \times 28}{24} = \frac{0.0448}{24} \approx 0.0018667$$ Sum all: $$ y_1 = 1 + 0.2 + 0.04 + 0.0106667 + 0.0018667 = 1.2525334 $$ 6. **Repeat for $$x_1=0.2$$, $$y_1=1.2525334$$:** Calculate derivatives at $$x=0.2$$, $$y=1.2525334$$: - $$y'_1 = (0.2)^2 + (1.2525334)^2 = 0.04 + 1.56984 = 1.60984$$ - $$y''_1 = 2(0.2) + 2(1.2525334)(1.60984) = 0.4 + 4.0319 = 4.4319$$ - $$y'''_1 = 2 + 2((1.60984)^2 + 1.2525334 \times 4.4319) = 2 + 2(2.592 + 5.553) = 2 + 2(8.145) = 2 + 16.29 = 18.29$$ - $$y^{(4)}_1 = 2(3 \times 1.60984 \times 4.4319 + 1.2525334 \times 18.29) = 2(21.4 + 22.9) = 2(44.3) = 88.6$$ 7. **Calculate $$y_2 = y(0.4)$$:** $$ y_2 = y_1 + h y'_1 + \frac{h^2}{2} y''_1 + \frac{h^3}{6} y'''_1 + \frac{h^4}{24} y^{(4)}_1 $$ Substitute values: $$ y_2 = 1.2525334 + 0.2 \times 1.60984 + \frac{0.04}{2} \times 4.4319 + \frac{0.008}{6} \times 18.29 + \frac{0.0016}{24} \times 88.6 $$ Calculate each term: - $$0.2 \times 1.60984 = 0.321968$$ - $$0.02 \times 4.4319 = 0.088638$$ - $$\frac{0.008 \times 18.29}{6} = \frac{0.14632}{6} = 0.024387$$ - $$\frac{0.0016 \times 88.6}{24} = \frac{0.14176}{24} = 0.005907$$ Sum all: $$ y_2 = 1.2525334 + 0.321968 + 0.088638 + 0.024387 + 0.005907 = 1.6934334 $$ **Final approximate values:** - $$y(0.2) \approx 1.2525$$ - $$y(0.4) \approx 1.6934$$ These are the solutions using the Taylor series method of order four with step size 0.2.