Chapter 1 - First-Order Differential Equations - 1.10 Numerical Solution to First-Order Differential Equations - Problems - Page 101: 8

$y(0.5)=1.0895$

We have $$y'=x-y^2$$ Setting $h=0.05$ in equation $y'=x-y^2$ $$y_{n+1}=y_n+0.025(x_n-y_n^2+x_{n+1}-y^2_{n+1})$$ Hence, $y_1=1.8209$ $y_2=1.6725$ $y_3=1.5497$ $y_4=1.4468$ $y_5=1.36$ $y_6=1.2866$ $y_7=1.2243$ $y_8=1.1715$ $y_9=1.1269$ $y_{10}=1.0895$ If we increase the step size to $h = 0.05$, the corresponding approximation to $y(0.5)$ becomes $y_{10}=1.0895$