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

Formula

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

$k_1=hf(x_0,y_0,z_0)$

$l_1=hg(x_0,y_0,z_0)$

$k_2=hf(x_0+h,y_0+k_1,z_0+l_1)$

$l_2=hg(x_0+h,y_0+k_1,z_0+l_1)$

$y_1=y_0+(k_1+k_2)/2$

Method-2 :

$k_1=hf(x_0,y_0,z_0)$

$l_1=hg(x_0,y_0,z_0)$

$k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)$

$l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)$

$y_1=y_0+k_2$

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

Solution:
Given

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

put

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

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

We have system of equations

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

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

Method-1 : Using formula $k_2=hf(x_0+h,y_0+k_1,z_0+l_1)$

Second order R-K method for second order differential equation

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

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

$k_2=hf(x_0+h,y_0+k_1,z_0+l_1)=(0.1)*f(0.1,1,0.1)=(0.1)*(0.1)=0.01$

$l_2=hg(x_0+h,y_0+k_1,z_0+l_1)=(0.1)*g(0.1,1,0.1)=(0.1)*(1.199)=0.1199$

$y_1=y_0+(k_1+k_2)/2=1+0.005=1.005$

$:.y(0.1)=1.005$

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.
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1. Formula-1 & Example-1