Develop a C-code to meet the requirements 1) Find the solution of the differenti

Question

Develop a C-code to meet the requirements 1) Find the solution of the differential equation below by using Euler's method. The range of x is 3 to 4 dy/dx-x2+2x with y-3 at x-3 2 Find the appropriate step size for this specific problem. 3) Print all the y values within the range of x-3 to x-4 4) Solve the differential equation analytically and compare with the Euler's method 5) Obtain a linear regression y-ax+b and show the r value Your C-code must contain the following features: (1) use of arrays, (2) use of pointers, (3) use of structure, (4) use of union, (5) use of functions and function calls, (6) formatted output on screen and (7) saving of the same output on a file You need to submit a written report on the project and the report needs to follow the format below ii Cover page (see attached as an ix) Summary (x) Introduction tells what you want to do in this project (xi Theory of Euler's method (xii) Results and Discussion (xiii Conclusion (xiv) Appendix a. List of your C code b. Sample Run of your Code

Explanation / Answer

CODE:

#include<stdio.h>

#include <math.h>

double sum(double [], double [], int n);

double r(double [], double [], int , double , double );

float func(float ,float );

float euler(float ,float ,float ,int );

int main(){

double x[]={2,3};

double y[]={3};

double one[]={1,1};

float xn,y0,e, x0=2;

int n=2;

double sumx, sumy,sumxx,sumxy;

double a,b;

printf(" inputn: ");

scanf("%d",&n);

printf("input x0:");

scanf("%f",&x0);

printf("input xn:");

scanf("%f",&xn);

printf("input y0: ");

scanf("%f",&y0);

sumx=sum(x,one,n);

sumy=sum(y,one,n);

sumxx=sum(x,x,n);

sumxy=sum(x,y,n);

a=(n*sumxy-sumx*sumy)/(n*sumxx-sumx*sumx);

b=(sumy*sumxx-sumx*sumxy)/(n*sumxx-sumx*sumx);

e = euler(x0,xn,y0,n);

printf(" By linear regression, the function is: y=%fx+%f ",a,b);

printf(" When x= %f, y= %f ",a*x0+b,x0);

printf(" r = %f ",r(x,y,n,a,b));

return 0;

}

double sum(double x[],double y[], int n){

double s=0.0;int i=0;

for (i=0;i<n;i++)

s=s+x[i]*y[i];

return s;

}

double r(double x[],double y[], int n, double a, double b){

double s0=0,s=0,ybar=0,total=0;

int i=0;

for (i=0;i<n;i++){

total+=y[i];

}

ybar=total/n;

for (i=0;i<n;i++){

s0+=pow((y[i]-ybar),2);

s+=pow((y[i]-b-a*x[i]),2);

}

return pow((s0-s)/s0,0.5);

}

float func(float x,float y){

return ((x*x)-(3*x));

}

float euler(float x0,float xn,float y0,int n){

float x,y,h;

int i;

x=x0;

y=y0;

h=(xn-x0)/n;

printf("y(%f)=%6.4f ",x0,y0);

for(i=1;i<=n;i++) {y=y+h*func(x,y);

x=x0+i*h;

printf("y(%f)=%6.4f ",x,y);

}

return y;

}

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