Answer
$y_1=2, y_2=2.0202,y_3=2.0618$, $y=e^{x^2}+1$, $y(0.3)=e^{(0.3)^2}+1 \approx 2.0942$
Work Step by Step
Use $y_{n+1}=y_n+f(x_n,y_n) dx$ , we have $x_0=0,x_1=0+0.1=0.1, x_2=0.1+0.1=0.2$ when $y(0)=2$
and $y_1=2, y_2=2.0202,y_3=2.0618$
Now, $\int dy=\int (2x) [e^{x^2}] dx$ and $y=e^{(x^2)}+c$
Next, apply the initial conditions $y(0)=2$ , we get
$ \implies e^{(0^2)}+c=2$ or, $c=1$
Now, $y=e^{x^2}+c \implies y=e^{x^2}+1$
Hence, $y(0.3)=e^{(0.3)^2}+1 \approx 2.0942$
