C語言 尤拉方法Euler Method 求一階常微分方程式ODEdy/dx=(x + y + xy) y(0.1) 的解

C語言 尤拉方法Euler Method 求一階常微分方程式ODEdy/dx=(x + y + xy)    y(0.1)   的解

源自於 https://www.geeksforgeeks.org/euler-method-solving-differential-equation/

Euler Method :
In mathematics and computational science, the Euler method (also called forward
Euler method) is a first-order numerical procedurefor solving ordinary differential
equations (ODEs) with a given initial value.
Consider a differential equation dy/dx = f(x, y) with initialcondition y(x0)=y0
then succesive approximation of this equation can be given by:

y(n+1) = y(n) + h * f(x(n), y(n))

where h = (x(n) – x(0)) / n
h indicates step size. Choosing smaller
values of h leads to more accurate results
and more computation time.

/* C Program to find approximation of a ordinary
   differential equation using euler method.*/

#include <stdio.h>
#include <math.h>
 
// Consider a differential equation
// dy/dx=(x + y + xy)
float func(float x, float y)
{
    return (x + y + x * y);
}
 
// Function for Euler formula
void euler(float x0, float y, float h, float x)
{
    float temp = -0;
 
    // Iterating till the point at which we
    // need approximation
    int i=0;
    printf("%d , Initial Values  y0 = 1 , x0 = 0; \n",i);
    i++;
    while (x0 < x) {
        temp = y;
        y = y + h * func(x0, y);
        // y1 = y0 + h * f(x0, y0)
        printf("%d , yn+1=yn + h* f(xn ,yn)= %0.4lf+ %0.4lf* %0.4lf= %0.4lf\n",i,temp,h,func(x0, temp),y );
        x0 = x0 + h;
        i++;
    }
 
    // Printing approximation
    printf("Approximate solution at x = %0.3lf is %0.6lf ",x, y);
   
}
 
// Driver program
int main()
{
    // Initial Values
    float x0 = 0;
    float y0 = 1;
    float h = 0.025;
 
    // Value of x at which we need approximation
    float x = 0.1;
 
    euler(x0, y0, h, x);
    return 0;
}

輸出畫面
0 , Initial Values  y0 = 1 , x0 = 0;
1 , yn+1=yn + h* f(xn ,yn)= 1.0000+ 0.0250* 1.0000= 1.0250
2 , yn+1=yn + h* f(xn ,yn)= 1.0250+ 0.0250* 1.0756= 1.0519
3 , yn+1=yn + h* f(xn ,yn)= 1.0519+ 0.0250* 1.1545= 1.0808
4 , yn+1=yn + h* f(xn ,yn)= 1.0808+ 0.0250* 1.2368= 1.1117
Approximate solution at x = 0.100 is 1.111673