Use Euler’s method with step size \[{\bf{h = }}\frac{{\bf{1}}}{{\bf{2}}}\] to approximate the solution to the initial value problem \[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = x - }}{{\bf{y}}^{\bf{2}}}{\bf{,}}\;{\bf{y(1) = 2}}\] at x = 2.

\[\begin{array}{l}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{x}}_{\bf{n}}}{\bf{ + h}}\\{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{y}}_{\bf{n}}}{\bf{ + h}}\;{\bf{f(x,y)}}\end{array}\]

For,

\[\begin{array}{c}{{\bf{x}}_{\bf{o}}}{\bf{ = 1,}}{{\bf{y}}_{\bf{o}}}{\bf{ = 2}}\\{{\bf{x}}_{\bf{1}}}{\bf{ = }}{{\bf{x}}_{\bf{o}}}{\bf{ + 0}}{\bf{.5 = 1 + 0}}{\bf{.5 = 1}}{\bf{.5}}\\{{\bf{x}}_{\bf{2}}}{\bf{ = }}{{\bf{x}}_{\bf{1}}}{\bf{ + 0}}{\bf{.5 = 1}}{\bf{.5 + 0}}{\bf{.5 = 2}}\\{{\bf{y}}_{\bf{1}}}{\bf{ = }}{{\bf{y}}_{\bf{o}}}{\bf{ + 0}}{\bf{.5(}}{{\bf{x}}_{\bf{o}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}_{\bf{o}}{\bf{) = 2 + 0}}{\bf{.5(1 - }}{{\bf{2}}^{\bf{2}}}{\bf{) = 0}}{\bf{.5}}\\{{\bf{y}}_{\bf{2}}}{\bf{ = 0}}{\bf{.5 + 0}}{\bf{.5(1}}{\bf{.5 - 0}}{\bf{.}}{{\bf{5}}^{\bf{2}}}{\bf{) = 1}}{\bf{.125}}\end{array}\]

Hence, the solution is for{h = 0.5 atx = 2 is 1.125.