Solution:

$\frac{d y}{d x}=x^{2} y-1 \quad x_{0}=0, y_{0}=1, h=0.1$

To find : $y(0.1)$

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

Taylor's series is:

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

$y(0.1)=1+0.1(-1)+0+\frac{(0.1)^{3}}{3 !}(2)$

$\therefore y(0.1)=0.9003$