Write a C++ program that can solve the following Boundary-Value Second Order Dif

Question

Write a C++ program that can solve the following Boundary-Value Second Order Differential Equation: y" = -2 sin(x) e^5-x. with y(0) = 0 and y(pi) = pi Your program must use the Finite Differences Method. You can select any step size you consider reasonable; proof solution will be decided on how well the calculated results match the analytical solution at the upper limit, that is, where x = pi. As part of the presentation of results, you must prepare an Excel plot of the calculated values and the "exact" values. (Show the "exact values" as a continuous line and the calculated points as small markers. The analytical solution is: y = sin(x) e^5-x + x

Explanation / Answer

#include <stdio.h>
#include <math.h>

#define SIZE 100

      double X[SIZE];
      double Y[3][SIZE], Z[3][SIZE];
      double cc,er,xl,x0,h,ya,yb,yex;
      int k,kl,l;
    
      double F(double x, double y,double z)
      {
        return z;
      }

      // z'= -2sin(x)e^5-x
      double G(double x,double y,double z)
      {
        double gg;
        gg = -2sin(x)e^5-x;
        if (l==2) gg += 5.0*x*x;
        return gg;
      }

      double FX(double x)
      {
        return 8.0/x+x*x*x*(x+2.0);
      }

    
      void Boundary_Value(double x,double y,double z,double h,double *x1,double *y1, double *z1) {
        double c1,c2,c3,c4,d1,d2,d3,d4,h2;
        c1=F(x,y,z);
        d1=G(x,y,z);
        h2=h/2.0;
        c2=F(x+h2,y+h2*c1,z+h2*d1);
        d2=G(x+h2,y+h2*c1,z+h2*d1);
        c3=F(x+h2,y+h2*c2,z+h2*d2);
        d3=G(x+h2,y+h2*c2,z+h2*d2);
        c4=F(x+h,y+h*c3,z+h*d3);
        d4=G(x+h,y+h*c3,z+h*d3);
        *x1=x+h;
        *y1=y+h*(c1+2.0*c2+2.0*c3+c4)/6.0;
        *z1=z+h*(d1+2.0*d2+2.0*d3+d4)/6.0;
      }

void main()
{

x0=1.0;     
xl=2.0;     
kl=20;      
h=(xl-x0)/kl;
ya=11.0;    
yb=36.0;   
X[0]=x0;

for (l=1; l<3; l++)
{
    Y[l][0]=(l-1)*ya; Z[l][0]=1.0;
    for (k=0; k<kl; k++)
      Boundary_Value(X[k],Y[l][k],Z[l][k],h,&X[k+1],&Y[l][k+1],&Z[l][k+1]);
}

}
cc=(yb-Y[2][kl])/Y[1][kl];
printf(" ");
printf("X Y estimated    Y true    Absolute Error ");
for (k=0; k<=kl; k++)
{
    Y[0][k]=Y[2][k] + cc*Y[1][k];
    yex=FX(X[k]);       
    er=fabs(yex-Y[0][k]);
    printf("%9.4f%12.7f %12.7f    %9.2e ", X[k],Y[0][k],yex,er);
}

}

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