Runge-Kutta 2 method (2nd order derivative) Formula-1 & Example-1

Formula

2. Second order R-K method
Method-1 :

k1=hf(x0,y0,z0) $k_1=hf(x_0,y_0,z_0)$

l1=hg(x0,y0,z0) $l_1=hg(x_0,y_0,z_0)$

k2=hf(x0+h,y0+k1,z0+l1) $k_2=hf(x_0+h,y_0+k_1,z_0+l_1)$

l2=hg(x0+h,y0+k1,z0+l1) $l_2=hg(x_0+h,y_0+k_1,z_0+l_1)$

y1=y0+k1+k22 $y_1=y_0+(k_1+k_2)/2$

Method-2 :

k1=hf(x0,y0,z0) $k_1=hf(x_0,y_0,z_0)$

l1=hg(x0,y0,z0) $l_1=hg(x_0,y_0,z_0)$

k2=hf(x0+h2,y0+k12,z0+l12) $k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)$

l2=hg(x0+h2,y0+k12,z0+l12) $l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)$

y1=y0+k2 $y_1=y_0+k_2$

Examples
1. Find y(0.1) for y′′=1+2xy-x2z $y''=1+2xy-x^2z$ , x0=0,y0=1,z0=0 $x_0=0, y_0=1, z_0=0$ , with step length 0.1 using Runge-Kutta 2 method (2nd order derivative)

Solution:
Given

y′′=1+2xy-x2z,y(0)=1,y′(0)=0,h=0.1,y(0.1)=? $y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, h=0.1, y(0.1)=?$

put

dydx=z $(dy)/(dx)=z$ and differentiate w.r.t. x, we obtain

d2ydx2=dzdx $(d^2y)/(dx^2)=(dz)/(dx)$

We have system of equations

dydx=z=f(x,y,z) $(dy)/(dx)=z=f(x,y,z)$

dzdx=1+2xy-x2z=g(x,y,z) $(dz)/(dx)=1+2xy-x^2z=g(x,y,z)$

Method-1 : Using formula k2=hf(x0+h,y0+k1,z0+l1) $k_2=hf(x_0+h,y_0+k_1,z_0+l_1)$

Second order R-K method for second order differential equation

k1=hf(x0,y0,z0)=(0.1)⋅f(0,1,0)=(0.1)⋅(0)=0 $k_1=hf(x_0,y_0,z_0)=(0.1)*f(0,1,0)=(0.1)*(0)=0$

l1=hg(x0,y0,z0)=(0.1)⋅g(0,1,0)=(0.1)⋅(1)=0.1 $l_1=hg(x_0,y_0,z_0)=(0.1)*g(0,1,0)=(0.1)*(1)=0.1$

k2=hf(x0+h,y0+k1,z0+l1)=(0.1)⋅f(0.1,1,0.1)=(0.1)⋅(0.1)=0.01 $k_2=hf(x_0+h,y_0+k_1,z_0+l_1)=(0.1)*f(0.1,1,0.1)=(0.1)*(0.1)=0.01$

l2=hg(x0+h,y0+k1,z0+l1)=(0.1)⋅g(0.1,1,0.1)=(0.1)⋅(1.199)=0.1199 $l_2=hg(x_0+h,y_0+k_1,z_0+l_1)=(0.1)*g(0.1,1,0.1)=(0.1)*(1.199)=0.1199$

y1=y0+k1+k22=1+0.005=1.005 $y_1=y_0+(k_1+k_2)/2=1+0.005=1.005$

$:.y(0.1)=1.005$

Method-2 : Using formula $k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)$

Second order R-K method for second order differential equation

$k_1=hf(x_0,y_0,z_0)=(0.1)*f(0,1,0)=(0.1)*(0)=0$

$l_1=hg(x_0,y_0,z_0)=(0.1)*g(0,1,0)=(0.1)*(1)=0.1$

$k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*f(0.05,1,0.05)=(0.1)*(0.05)=0.005$

$l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*g(0.05,1,0.05)=(0.1)*(1.09988)=0.10999$

$y_1=y_0+k_2=1+0.005=1.005$

$:.y(0.1)=1.005$

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here

1. Formula-1 & Example-1