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PLEASE HELP ME FIGURE OUT THIS CODE i really need help with this. I need help fi

ID: 3702764 • Letter: P

Question

PLEASE HELP ME FIGURE OUT THIS CODE i really need help with this. I need help figuring out how to solve the errors.

This is my code and need help revising it. PLEASE just revise my code.

#include<stdio.h>

#include <math.h>

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

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

float func(float x,float y);

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

int main(){

  

double x[]={2,3};

double y[]={3};

double one[]={1,1};

  

float x0,xn,y0,e;

int n=2;

  

double x0=2;

double sumx, sumy,sumxx,sumxy;

double a,b;

  

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);

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

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

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);

}

printf(" inputn: ");

scanf("%d",&n);

printf("input x0: ");

scanf("%f",&x0);

printf("input xn: ");

scanf("%f",&xn);

printf("input y0: ");

scanf("%f",&y0);

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

}

}

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;

}

Develop a C-code to meet the requirements below: 1) Find the solution of the differential equation below by using Euler's method. The range of x is 2 to 3. dyldx-x2-3x with y-3 at x-2 Find the appropriate step size for this specific problem 3) Print all the y values within the range of x-2 to x-3. 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

Explanation / Answer

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


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

/*initialization should be done like this. should not specify the variable names in the function declaration part*/

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,yo,e;

    int n=2;

    double x0=2;

    double sumx, sumy,sumxx,sumxy;

    double a,b;

    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);

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

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

    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);

    }


/*   printf(" inputn: ");

    scanf("%d",&n);

    printf("input x0:");            here this code is useless. if you want the read the x0 n xn values from the user dynamically please use this code in main function without reading static data use this code then it will work.

    scanf("%f",&x0);

    printf("input xn:");

    scanf("%f",&xn);

    printf("input y0: ");

    scanf("%f",&y0);


    e=euler(x0,xn,y0,n);

}

}*/

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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