Как линии трендов на диаграммах в принятии решений?

    #include <math.h>
    #include <stdio.h>
double y1(double x,double y) {
    return (2*y-sqrt(x*x+y*y))/(2*x);
};

double ye(double x) {
    return (sinh(0.5*log(1/x)));
};

int main() {
    double y0=0, x0=1, xn=2, h=0.05;
    double x,y,yr,et;
    x=x0; y=y0;
    for (int i=1; i<=floor((xn-x0)/h)+1; i++) {
        yr=y+h*y1(x,y);
        et=ye(x);
        printf("\nx=%8.5f, y=%8.5f, yr=%8.5f, et=%8.5f, eps=%8.5f",x,y,yr,et,yr-et);
        x+=h; y=yr;
    };
    return 0;   
}

Тестовое решение:
x= 1.00000, y= 0.00000, yr=-0.02500, et= 0.00000, eps=-0.02500
x= 1.05000, y=-0.02500, yr=-0.05120, et=-0.02440, eps=-0.02680
x= 1.10000, y=-0.05120, yr=-0.07855, et=-0.04767, eps=-0.03088
x= 1.15000, y=-0.07855, yr=-0.10703, et=-0.06994, eps=-0.03709
x= 1.20000, y=-0.10703, yr=-0.13658, et=-0.09129, eps=-0.04530
x= 1.25000, y=-0.13658, yr=-0.16720, et=-0.11180, eps=-0.05539
x= 1.30000, y=-0.16720, yr=-0.19883, et=-0.13156, eps=-0.06727
x= 1.35000, y=-0.19883, yr=-0.23147, et=-0.15062, eps=-0.08085
x= 1.40000, y=-0.23147, yr=-0.26507, et=-0.16903, eps=-0.09604
x= 1.45000, y=-0.26507, yr=-0.29963, et=-0.18685, eps=-0.11277
x= 1.50000, y=-0.29963, yr=-0.33511, et=-0.20412, eps=-0.13098
x= 1.55000, y=-0.33511, yr=-0.37150, et=-0.22089, eps=-0.15061
x= 1.60000, y=-0.37150, yr=-0.40877, et=-0.23717, eps=-0.17160
x= 1.65000, y=-0.40877, yr=-0.44691, et=-0.25301, eps=-0.19390
x= 1.70000, y=-0.44691, yr=-0.48591, et=-0.26844, eps=-0.21747
x= 1.75000, y=-0.48591, yr=-0.52574, et=-0.28347, eps=-0.24226
x= 1.80000, y=-0.52574, yr=-0.56638, et=-0.29814, eps=-0.26824
x= 1.85000, y=-0.56638, yr=-0.60784, et=-0.31247, eps=-0.29537
x= 1.90000, y=-0.60784, yr=-0.65008, et=-0.32646, eps=-0.32362
x= 1.95000, y=-0.65008, yr=-0.69310, et=-0.34015, eps=-0.35295
x= 2.00000, y=-0.69310, yr=-0.73689, et=-0.35355, eps=-0.38334