(1)

$\begin{aligned} \frac{d y}{d x} &=x y+1 \quad, x_{0}=0, y_{0}=0, \mathrm{h}=0.1 \\ f(x, y) &=1+x y \end{aligned}$

$\begin{array}{ll}{y^{\prime}=1+x y} & {y_{0}^{\prime}=1} \\ {y^{\prime \prime}=x y^{\prime}+y} & {y_{0}^{\prime \prime}=0} \\ {y^{\prime \prime \prime}=x y^{\prime \prime}+2 y^{\prime}} & {y_{0}^{\prime \prime \prime}=2}\end{array}$

Taylor's series is given by

$\quad y(0.1)=y_{0}+h . y_{0}^{\prime}+\frac{h^{2}}{2 !} y_{0}^{\prime \prime}+\frac{h^{3}}{3 !} y_{0}^{\prime \prime \prime}+\cdots$

$\begin{aligned} &=0+0.1(1)+0+\frac{(0.1)^{3}}{6}(2) \\ y(0.1) &=0.1003 \end{aligned}$

(2)

$x_{1}=0.1, y_{1}=0.1003, h=0.1$

$\begin{array}{rl}{y^{\prime}=1+x y}, & {y_{0}^{\prime}=1.01003} \\ {y^{\prime \prime}=x y^{\prime}+y}, & {y_{0}^{\prime \prime}=0.201303} \\ {y^{\prime \prime \prime}=x y^{\prime \prime}+2 y^{\prime}}, & {y_{0}^{\prime \prime \prime}=2.0401903}\end{array}$

$\therefore \quad y(0.2)=0.1003+1.01003(0.1)+\frac{0.1^{2}}{2 !}(0.201303)+\frac{0.1^{3}}{6}(2.0401903)$

$\therefore y(0.2)=0.202708$