Answer
The solution is $y_{10}=1.044$
Work Step by Step
We have
$$y'=x-y^2$$
Setting $h=0.05$ in equation $y'=x-y^2$
$$y_{n+1}=y_n+0.05(x_n-y_n^2)$$
Hence,
$y_1=y_0+0.05(x_0-y_0^2)=2+0.05(0-2^2)=1.8$
$y_2=y_1+0.05(x_1-y_1^2)=1.8+0.05(0.05-1.8^2)=1.6405$
$y_3=y_2+0.05(x_2-y_2^2)=1.6405+0.05(0.1-1.6405^2)=1.51$
$y_4=y_3+0.05(x_3-y_3^2)=1.51+0.05(0.15-1.51^2)=1.403$
$y_5=y_4+0.05(x_4-y_4^2)=1.403+0.05(0.2-1.403^2)=1.314$
$y_6=y_5+0.05(x_5-y_5^2)=1.314+0.05(0.25-1.314^2)=1.24$
$y_7=y_6+0.05(x_6-y_6^2)=1.24+0.05(0.3-1.24^2)=1.178$
$y_8=y_7+0.05(x_7-y_7^2)=1.178+0.05(0.35-1.178^2)=1.126$
$y_9=y_8+0.05(x_8-y_8^2)=1.126+0.05(0.4-1.126^2)=1.08$
$y_{10}=y_9+0.05(x_9-y_9^2)=1.08+0.05(0.45-1.08^2)=1.044$
If we increase the step size to $h = 0.05$, the corresponding approximation to $y(0.5)$ becomes $y_{10}=1.044$
