Solution:

The Taylor series is given by,

$y=y_{0}+x y_{0}^{\prime}+\frac{x^{2}}{2 !} y_{0}^{\prime \prime}+\frac{x^{3}}{3 !} y_{0}^{\prime \prime \prime} \ldots . .(1)$

With $x_{0}=0, y_{0}=0.2, x=0.4$

$\begin{aligned} \text { Now, } y^{\prime} &=1+x y & \therefore y_{0}^{\prime}=1 \\ y^{\prime \prime} &=y+x y & \therefore y_{0}^{\prime \prime}=y_{0}=0.2 \end{aligned}$

$\begin{aligned} \mathrm{y}^{\prime \prime \prime} &=\mathrm{y}^{\prime}+\mathrm{y}+\mathrm{xy}^{\prime \prime} \\ &=2 \mathrm{y}+\mathrm{xy}^{\prime \prime} \quad \therefore y_{0}^{\prime \prime \prime}=2 y_{0}^{\prime}=2 \end{aligned}$

$y^{\prime \prime \prime}=2 y^{\prime \prime}+y^{\prime \prime}+x y^{\prime \prime} \quad \therefore y_{0}^{\prime \prime \prime}=3 y_{0}^{\prime \prime}+x y_{0}^{\prime \prime \prime}=0.6$

Putting these values in equation $1,$ we get

$y=0.2+(0.4) 1+\frac{(0.4)^{2}}{2 !} 0.2+\frac{(0.4)^{3}}{3 !} .2+\frac{(0.4)^{4}}{4 !} \cdot(0.6)+\cdots$

$y=0.2+0.4+0.016+0.02133+0.00064$

$y=0.63797$