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TNPL JoongJin-Cho Runge-kutta’s method of order 4

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TNPL JoongJin-Cho This routine solve the initial value problem at equidistant points Here the function f(x,t) is continuous the at equidistant points Here the function f(x,t) is continuous theinterval Algorithm

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TNPL JoongJin-Cho Algorithm

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TNPL JoongJin-Cho Runge-Kutta ’ s method of order 4 to solve a first-order differential equation Algorithm

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TNPL JoongJin-Cho public static void main(String args[]){ int i; double t,ti,h,x,xi, N=5; double f; //Differential equation dx/dt=f(x,t)=x+t double t0,x0,f1,f2,f3,f4; x=xi; t=ti; f=x+t; for(i=0;i

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TNPL JoongJin-Cho import java.io.*; public class RungeKutta2{ static BufferedReader bf = new BufferedReader (new InputStreamReader(System.in)); //4th order Runge-Kutta method, need derivative f public static void main(String args[])throws IOException { int i; double t,ti,h,x,xi,N=5; double f; double t0,x0,f1,f2,f3,f4; String str; System.out.print("Input initial value ti = >"); str=bf.readLine(); ti=Double.parseDouble(str); System.out.print("Input initial value xi = >"); str=bf.readLine(); xi=Double.parseDouble(str); System.out.print("step size h = >"); str=bf.readLine(); h=Double.parseDouble(str);

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TNPL JoongJin-Cho x=xi; t=ti; f=x+t; for(i=0;i

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1 Convert this problem into our standard form: Minimize 3 x 1 + 4.5 x 2 – 6 x 3 subject to 1 x 1 - 2 x 2 – 3 x 3 ≥ 6.2 4 x 1 + 5 x 2 – 6 x 3 ≤ 5 7 x 1.

1 Convert this problem into our standard form: Minimize 3 x 1 + 4.5 x 2 – 6 x 3 subject to 1 x 1 - 2 x 2 – 3 x 3 ≥ 6.2 4 x 1 + 5 x 2 – 6 x 3 ≤ 5 7 x 1.

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