Answer
$y(0.6) \approx 0.0603$
Work Step by Step
We are given $\frac{dy}{dx}=x+y^{2}$
so that $g(x,y)=x+y^{2}$
Since $x=0, y=0$
$g(x_{0},y_{0})=0$
and $y_{1}=y_{0}+g(x_{0},y_{0})h=0+0\times0.1=0$
Now $x_{1}=0.1, y_{1}=0$ and $g(x_{1},y_{1})=0.1+0^{2}=0.1$
Then $y_{2}=0+0.1\times0.1=0.01$
$y_{3}=0+(0.2001)\times0.1=0.02001$
$y_{4}=0+(0.3004)\times0.1=0.03004$
$y_{5}=0+(0.4009)\times0.1=0.04009$
$y_{6}=0+(0.5016)\times0.1=0.05016$
$y_{7}=0+(0.6025)\times0.1=0.06025$
$y(0.6)$ mean $x=0.6 \rightarrow y_{7}=0.06025 \approx 0.0603$
