dy dx = y - 2x y = f x , y Now, y( 0 ) = 1 ⇒ y0 = 1 , x0 = 0 By Euler’s method : yn+1 = yn + h [ f ( xn , yn ) ] Where, xn = x0 + n h , n = 0, 1, 2, 3, … For, n = 0 For, x1 = x0 + h = 0 + 0.1 = 0.1 y1 = y0 + h f ( x0 , y0 ) y1 = 1 + ( 0.1 ) f ( 0 , 1 ) y1 = 1 + ( 0.1 ) 1 - 2 ( 0 )1 y1 = 1.1 For, n=1 Now, x2 = x0 + 2h = 0 + 0.2 = 0.2 y2 = y1 + h f ( x1 , y1 ) y1 = 1.1 + ( 0.1 ) f ( 0.1 , 1.1 ) y1 = 1 + ( 0.1 ) 1.1 - 2 ( 0.1 )1.1 y1 = 1.9182