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
#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 hint n = (int)((x - x0) / h);float k1, k2, k3, k4, k5;// Iterate for number of iterationsfloat y = y0;for (int i=1; i<=n; i++){// Apply Runge Kutta Formulas to find// next value of yk1 = 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 yy = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;// Update next value of xx0 = x0 + h;}return y;}// Driver Codeint 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：
// 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 hint n = (int)((x - x0) / h);float k1, k2, k3, k4, k5;// Iterate for number of iterationsfloat y = y0;for (int i=1; i<=n; i++){// Apply Runge Kutta Formulas to find// next value of yk1 = 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 yy = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;// Update next value of xx0 = x0 + h;}return y;}// Driver methodint 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：
// Java program to implement Runge Kutta methodimport 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 hint n = (int)((x - x0) / h);double k1, k2, k3, k4, k5;// Iterate for number of iterationsdouble y = y0;for (int i = 1; i <= n; i++){// Apply Runge Kutta Formulas to find// next value of yk1 = 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 yy = y + (1.0 / 6.0) * (k1 + 2 * k2 + 2 * k3 + k4);// Update next value of xx0 = 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
# 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 hn = (int)((x - x0)/h)# Iterate for number of iterationsy = y0for 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 yy = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4)# Update next value of xx0 = x0 + hreturn y# Driver methodx0 = 0y = 1x = 2h = 0.2print('Thevalueofyatxis:',rungeKutta(x0,y,x,h))
The value of y at x is: 1.1036393232374955
