Answer
$y(1)=0.994$
Work Step by Step
We are given:
$y'=2xy^2$
Setting $h=0.1$ in equation $y'=2xy^2$
we have:
$y_{n+1}=y_n+0.05(x_ny_n^2+x_{n+1}y_{n+1}^2)$
Hence,
$x_1=0.1 \rightarrow y_1=0.5025$
$x_2=0.2 \rightarrow y_2=0.5102$
$x_3=0.3 \rightarrow y_3=0.5235$
$x_4=0.4 \rightarrow y_4=0.5434$
$x_5=0.5 \rightarrow y_5=0.5713$
$x_6=0.6 \rightarrow y_6=0.6095$
$x_7=0.7 \rightarrow y_7=0.6617$
$x_8=0.8 \rightarrow y_8=0.7342$
$x_9=0.9 \rightarrow y_9=0.8379$
$x_{10}=1 \rightarrow y_{10}=0.994$
If we increase the step size to $h=0.1$, the corresponding approximation to y(1) becomes $y_{10}=0.994$
