Chapter 6 - Differential Equations - 6.1 Exercises - Page 405: 74

\begin{array}{|c|c|} \hline y_{1}&2.1\\ \hline y_{2}&2.2075 ,\\ \hline y_{3}&2.322875 \\ \hline y_{4}& 2.44651875 \\ \hline y_{5}&2.578844688 \\ \hline y_{6}&2.720286922 \\ \hline y_{7}&2.871301268 \\ \hline y_{8}&3.032366331 \\ \hline y_{9}&3.203984648 \\ \hline y_{10}&3.38668388 \\ \hline y_{11}&3.581018074 \\ \hline y_{12}&3.787568978 \\ \hline y_{13}&4.006947427 \\ \hline y_{14}&4.239794798 \\ \hline y_{15}&4.486784538 \\ \hline y_{16}&4.748623765 \\ \hline y_{17}&5.026054953 \\ \hline y_{18}&5.319857701 \\ \hline y_{19}&5.630850586 \\ \hline y_{20}&5.959893115 \\ \hline \end{array}

Given $$y^{\prime}=x+y, \quad y(0)=2, \ \ \ \ \ n=20, \quad h=0.05$$ So Using $h=0.05, x_{0}=0, y_{0}=2, \ \ $ $n=20$ and $F(x, y)=x-y,$ Use Euler's method in the form $$ x_{n}= x_{{n-1}}+ h$$ $$ y_{n}= y_{n-1}+ h F(x_{n-1},y_{n-1})$$ we get \begin{array}{|c|} \hline y_{1}=y_{0}+h F\left(x_{0}, y_{0}\right)=2+(0.05)(0+2)=2.1\\ \hline y_{2}=y_{1}+h F\left(x_{1}, y_{1}\right)=2.1+(0.05)(0.05.05+2.1)=2.2075 ,\\ \hline y_{3}=y_{2}+h F\left(x_{2}, y_{2}\right)=2.2075+(0.05)(0.1+2.2075)=2.322875 \\ \hline y_{4}=y_{3}+h F\left(x_{3}, y_{3}\right)=2.322875 +(0.05)(0.15+2.322875 )=2.44651875 \\ \hline y_{5}=y_{4}+h F\left(x_{4}, y_{4}\right)=2.44651875 +(0.05)(0.2+2.44651875 )=2.578844688 \\ \hline y_{6}=y_{5}+h F\left(x_{5}, y_{5}\right)=2.578844688 +(0.05)(0.25+2.578844688 )=2.720286922 \\ \hline y_{7}=y_{6}+h F\left(x_{6}, y_{6}\right)=2.720286922 +(0.05)(0.3+2.720286922 )=2.871301268 \\ \hline y_{8}=y_{7}+h F\left(x_{7}, y_{7}\right)=2.871301268 +(0.05)(0.35+2.871301268 )=3.032366331 \\ \hline y_{9}=y_{8}+h F\left(x_{8}, y_{8}\right)=3.032366331 +(0.05)(0.4+3.032366331 )=3.203984648 \\ \hline y_{10}=y_{9}+h F\left(x_{9}, y_{9}\right)=3.203984648 +(0.05)(0.45+3.203984648 )=3.38668388 \\ \hline y_{11}=y_{10}+h F\left(x_{10}, y_{10}\right)=3.38668388 +(0.05)(0.5+3.38668388 )=3.581018074 \\ \hline y_{12}=y_{11}+h F\left(x_{11}, y_{11}\right)=3.581018074 +(0.05)(0.55+3.581018074 )=3.787568978 \\ \hline y_{13}=y_{12}+h F\left(x_{12}, y_{12}\right)=3.787568978 +(0.05)(0.6+3.787568978 )=4.006947427 \\ \hline y_{14}=y_{13}+h F\left(x_{13}, y_{13}\right)=4.006947427 +(0.05)(0.65+4.006947427 )=4.239794798 \\ \hline y_{15}=y_{14}+h F\left(x_{14}, y_{14}\right)=4.239794798 +(0.05)(0.7+4.239794798 )=4.486784538 \\ \hline y_{16}=y_{15}+h F\left(x_{15}, y_{15}\right)=4.486784538 +(0.05)(0.75+4.486784538 )=4.748623765 \\ \hline y_{17}=y_{16}+h F\left(x_{16}, y_{16}\right)=4.748623765 +(0.05)(0.8+4.748623765 )=5.026054953 \\ \hline y_{18}=y_{17}+h F\left(x_{17}, y_{17}\right)=5.026054953 +(0.05)(0.85+5.026054953 )=5.319857701 \\ \hline y_{19}=y_{18}+h F\left(x_{18}, y_{18}\right)=5.319857701 +(0.05)(0.9+5.319857701 )=5.630850586 \\ \hline y_{20}=y_{19}+h F\left(x_{19}, y_{19}\right)=5.630850586 +(0.05)(0.9+5.630850586 )=5.959893115 \\ \hline \end{array}