Chapter 1 - First-Order Differential Equations - 1.10 Numerical Solution to First-Order Differential Equations - Problems - Page 101: 4

The solution is $y_{5}=0.776$

We have $$y'=-x^2y$$ Setting $h=0.2$ in equation $y'=-x^2y$ $$y_{n+1}=y_n+0.2(-x_n^2y_n)$$ Hence, $y_1=y_0+0.2(-x_0^2y_0)=1+0.2(-0^2\times1)=1$ $y_2=y_1+0.2-x_1^2y_1)=1+0.2(-0.2^2\times1)=0.992$ $y_3=y_2+0.2(x_2^2y_2)=0.992+0.2(-0.4^2\times0.992)=0.96$ $y_4=y_3+0.2(-x_3^2y_3)=0.96+0.2(-0.6^2\times0.96)=0.89$ $y_5=y_4+0.2(-x_4^2y_4)=0.89+0.2(-0.8^2\times0.89)=0.776$ If we increase the step size to $h = 0.2$, the corresponding approximation to $y(1)$ becomes $y_{5}=0.776$