Answer
$y(1)=0.858$
Work Step by Step
We have
$$y'=2xy^2$$
Setting $h=0.1$ in equation $y'=2xy^2$
$$y_{n+1}=y_n+0.1(2xy^2)$$
Hence,
$y_1=y_0+0.1(2xy^2)=0.5$
$y_2=y_1+0.1(2xy^2)=0.505$
$y_3=y_2+0.1(2xy^2)=0.515$
$y_4=y_3+0.1(2xy^2)=0.531$
$y_5=y_4+0.1(2xy^2)=0.554$
$y_6=y_5+0.1(2xy^2)=0.584$
$y_7=y_6+0.1(2xy^2)=0.625$
$y_8=y_7+0.1(2xy^2)=0.68$
$y_9=y_8+0.1(2xy^2)=0.754$
$y_{10}=y_9+0.1(2xy^2)=0.858$
If we increase the step size to $h = 0.1$, the corresponding approximation to $y(1)$ becomes $y_{10}=0.858$
