Chapter 1 - First-Order Differential Equations - 1.10 Numerical Solution to First-Order Differential Equations - Problems - Page 101: 7

$y(1)=0.5012$

We have $$y'=-\frac{2xy}{1+x^2}$$ Setting $h=0.1$ in equation $y'=-\frac{2xy}{1+x^2}$ $$y_{n+1}=y_n+0.05(-\frac{2x_ny_n}{1+x_n^2}-\frac{2x_{n+1}y^*{n+1}}{1+x_{n+1}^2})$$ Hence, $y_1=0.99$ $y_2=0.9616$ $y_3=0.9177$ $y_4=0.8625$ $y_5=0.8007$ $y_6=0.7163$ $y_7=0.6721$ $y_8=0.6108$ $y_9=0.5536$ $y_{10}=0.5012$ If we increase the step size to $h = 0.1$, the corresponding approximation to $y(1)$ becomes $y_{10}=0.5012$