Chapter 10 - Differential Equations - Chapter Review - Review Exercises - Page 562: 48

$y(1)=2.433$

We are given $\frac{dy}{dx}=x+y^{-1}$ so that $g(x,y)=x+y^{-1}$ Since $x=0, y=1$ $g(x,y)=0+1^{-1}=1$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=1+1\times0.2=1.2$ Now $x_{1}=0.2, y_{1}=1.2$ and $g(x_{1},y_{1})=1.033$ Then $y_{2}=1.2+1.033\times0.2=1.407$ $y_{3}=1.407+(1.11)\times0.2=1.629$ $y_{4}=1.629+(1.214)\times0.2=1.872$ $y_{5}=1.872+(1.334)\times0.2=2.139$ $y_{6}=2.139+(1.468)\times0.2=2.433$ $y(1)=y_{6}=2.433$