取K=1K1 1 +-k,+-K 6 则(4式化为 h yn=yn1+(K1+4K2+K3) 1 1 K2=f(n-1+on-1+Ki K3=f(xn-1+h,yn-1+h(2K2-K1) yo=y(ro) (7)式称为三阶 Runge-Kutt法取 1 2 3 6 1 6 4 6 1 K = K + K + K 则(4)式化为 ( 4 ) 6 1 K1 K2 K3 h y y n = n- + + + ( , ) 1 = n-1 n-1 K f x y ) 2 , 2 ( 2 1 1 K1 h y h K f x = n- + n- + ( , (2 )) 3 1 1 K2 K1 K f x h y h = n- + n- + - -----------(7) ( ) 0 0 y = y x ï ï î ï ï í ì (7)式称为三阶Runge-Kutta方法