前期是分享了matlab下面实现四阶龙格库塔（Runge-Kutta）求解微分方程，这期分享一下C 、C、Java、Python下面的四阶龙格库塔（Runge-Kutta）求解微分方程。
前文传送门：matlab代码实现四阶龙格库塔求解微分方程
C 方法
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
// "dy/dx = (x - y)/2"
float dydx(float x, float y)
{
return((x - y)/2);
}
// Finds value of y for a given x using step size h
// and initial value y0 at x0.
float rungeKutta(float x0, float y0, float x, float h)
{
// Count number of iterations using step size or
// step height h
int n = (int)((x - x0) / h);
float k1, k2, k3, k4, k5;
// Iterate for number of iterations
float y = y0;
for (int i=1; i<=n; i )
{
// Apply Runge Kutta Formulas to find
// next value of y
k1 = h*dydx(x0, y);
k2 = h*dydx(x0 0.5*h, y 0.5*k1);
k3 = h*dydx(x0 0.5*h, y 0.5*k2);
k4 = h*dydx(x0 h, y k3);
// Update next value of y
y = y (1.0/6.0)*(k1 2*k2 2*k3 k4);;
// Update next value of x
x0 = x0 h;
}
return y;
}
// Driver Code
int main()
{
float x0 = 0, y = 1, x = 2, h = 0.2;
cout << "The value of y at x is : " <<
rungeKutta(x0, y, x, h);
return 0;
}C：
代码语言：javascript复制// C program to implement Runge Kutta method
#include<stdio.h>
// A sample differential equation "dy/dx = (x - y)/2"
float dydx(float x, float y)
{
return((x - y)/2);
}
// Finds value of y for a given x using step size h
// and initial value y0 at x0.
float rungeKutta(float x0, float y0, float x, float h)
{
// Count number of iterations using step size or
// step height h
int n = (int)((x - x0) / h);
float k1, k2, k3, k4, k5;
// Iterate for number of iterations
float y = y0;
for (int i=1; i<=n; i )
{
// Apply Runge Kutta Formulas to find
// next value of y
k1 = h*dydx(x0, y);
k2 = h*dydx(x0 0.5*h, y 0.5*k1);
k3 = h*dydx(x0 0.5*h, y 0.5*k2);
k4 = h*dydx(x0 h, y k3);
// Update next value of y
y = y (1.0/6.0)*(k1 2*k2 2*k3 k4);;
// Update next value of x
x0 = x0 h;
}
return y;
}
// Driver method
int main()
{
float x0 = 0, y = 1, x = 2, h = 0.2;
printf("nThe value of y at x is : %f",
rungeKutta(x0, y, x, h));
return 0;
}
Java：
代码语言：javascript复制// Java program to implement Runge Kutta method
import java.io.*;
class differential
{
double dydx(double x, double y)
{
return ((x - y) / 2);
}
// Finds value of y for a given x using step size h
// and initial value y0 at x0.
double rungeKutta(double x0, double y0, double x, double h)
{
differential d1 = new differential();
// Count number of iterations using step size or
// step height h
int n = (int)((x - x0) / h);
double k1, k2, k3, k4, k5;
// Iterate for number of iterations
double y = y0;
for (int i = 1; i <= n; i )
{
// Apply Runge Kutta Formulas to find
// next value of y
k1 = h * (d1.dydx(x0, y));
k2 = h * (d1.dydx(x0 0.5 * h, y 0.5 * k1));
k3 = h * (d1.dydx(x0 0.5 * h, y 0.5 * k2));
k4 = h * (d1.dydx(x0 h, y k3));
// Update next value of y
y = y (1.0 / 6.0) * (k1 2 * k2 2 * k3 k4);
// Update next value of x
x0 = x0 h;
}
return y;
}
public static void main(String args[])
{
differential d2 = new differential();
double x0 = 0, y = 1, x = 2, h = 0.2;
System.out.println("nThe value of y at x is : "
d2.rungeKutta(x0, y, x, h));
}
}Python3
代码语言：javascript复制# Python program to implement Runge Kutta method
# A sample differential equation "dy / dx = (x - y)/2"
def dydx(x, y):
return ((x - y)/2)
# Finds value of y for a given x using step size h
# and initial value y0 at x0.
def rungeKutta(x0, y0, x, h):
# Count number of iterations using step size or
# step height h
n = (int)((x - x0)/h)
# Iterate for number of iterations
y = y0
for i in range(1, n 1):
"Apply Runge Kutta Formulas to find next value of y"
k1 = h * dydx(x0, y)
k2 = h * dydx(x0 0.5 * h, y 0.5 * k1)
k3 = h * dydx(x0 0.5 * h, y 0.5 * k2)
k4 = h * dydx(x0 h, y k3)
# Update next value of y
y = y (1.0 / 6.0)*(k1 2 * k2 2 * k3 k4)
# Update next value of x
x0 = x0 h
return y
# Driver method
x0 = 0
y = 1
x = 2
h = 0.2
print ('The value of y at x is:', rungeKutta(x0, y, x, h))The value of y at x is: 1.1036393232374955
