int_sine_rk2

# int_sine_rk2 - yin[0 = y[i rk2(x[i yin yout h do one...

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/* int_sine_rk2.c * Example code to integrate sin(x) from zero to pi by solving an ODE. */ #include <stdio.h> #include <math.h> #define N 1 /* number of ODEs */ #define NSTEP 10 /* number of integration steps */ #define XMIN 0.0 /* starting point for integration */ #define XMAX 3.14159265 /* stopping point for integration */ #define Y0 0.0 /* initial value */ void rk2(float xin, float yin[], float yout[], float h); void derivs(float xin, float yin[], float dydx[]); int main() { int i=0; float h = (XMAX - XMIN)/(1.*NSTEP); /* stepsize for integration */ float yin[N], yout[N]; float x[NSTEP+1]; float y[NSTEP+1]; for (i = 0; i <= NSTEP; i++) /* Define array of x values */ x[i] = XMIN + h*i; y[0]=Y0; /* initial value */ for (i = 0; i < NSTEP; i++) /* do integration */

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Unformatted text preview: { yin[0] = y[i]; rk2(x[i], yin, yout, h); /* do one integration step */ y[i+1] = yout[0]; } printf("Estimate of integral = %f\n", y[NSTEP]); return 0; } void derivs(float xin, float yin, float dydx) { /* evaluate RHS of ODE */ dydx[0] = sin(xin); return; } void rk2(float xin, float yin, float yout, float h) { /* Second order Runge-Kutta scheme */ int i; float k1[N], k2[N], yt[N], dydx[N]; /* N is the number of ODEs */ /* Evaluate k1 */ derivs(xin, yin, dydx); for (i = 0; i < N; i++) { k1[i] = h*dydx[i]; yt[i] = yin[i] + 0.5*k1[i]; } /* Evaluate k2, then update the dependent variable */ derivs(xin + 0.5*h, yt, dydx); for (i = 0; i < N; i++) { k2[i] = h*dydx[i]; yout[i] = yin[i] + k2[i]; } return; }...
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## This note was uploaded on 09/29/2009 for the course COSC 1002 taught by Professor Wheatland during the Three '09 term at University of Sydney.

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int_sine_rk2 - yin[0 = y[i rk2(x[i yin yout h do one...

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