Har bir qiymati uchun takrorlaymiz. (6)-ustunni qiymatlarining yig’indisini hisoblab, natijani 6 ga bo’lamiz va

I-IV jarayonni Kⁱ ni (i=0,1,2, ...n) har bir qiymati uchun takrorlaymiz. (6)-ustunni qiymatlarining yig’indisini hisoblab, natijani 6 ga bo’lamiz va u=(1/6) (K₁⁽ⁱ⁾+2 K₂⁽ⁱ⁾+2 K₃⁽ⁱ⁾+ K₄⁽ⁱ⁾) ni
topamiz. Va nihoyat y_i+1=y_i+ y_i topiladi. YUqorida keltirilgan hisoblash tartibini [a,b] kesmani barcha nuqtalari uchun takrorlaymiz.
1-Jadval

	X	U	u’=f(x,y)	K=hf(x,y)	u
1	2	3	4	5	6
	x0	y0	f(x0 ,y0)	K1^(0)	K1^(0)
	x0+h/2	y0+K1^(0)/2	f(x0+h/2; y0+K1^(0)/2)	K2^(0)	2K2^(0)
0	x0+h/2	y0+K2^(0)/2	f(x0+h/2; y0+K2^(0)/2)	K3^(0)	2K3^(0)
	x0+h	y0+K3^(0)	f(x0+h; y0+K3^(0))	K4^(0)	K4^(0)
	x1	y1=y0+ y0	f(x1 ,y1)	K1^(0)	K1^(0)
	x1+h/2	y1+K1^(1)/2	f(x1+h/2; y1+K1^(1)/2)	K2^(0)	2K2^(0)
1	x1+h/2	y1+K2^(1)/2	f(x1+h/2; y1+K2^(1)/2)	K3^(0)	2K3^(0)
	x1+h	y1+K3^(1)	f(x1+h; y1+K3^(1))	K4^(0)	K4^(0)
2	x2	y2=y1+ y1

Misol. Runge-Kutta usuli yordamida quyidagi differensial tenglamaga qo’yilgan boshlang’ich masalaning
y^’= , u(1)=0 yechimi [1;1,5] kesmada h=0,1 qadam bilan