Use Euler's method to obtain a four-decimal approximation of the indicated value. Carry out the recursion of (3) in Section 2.6

Answer:

At h=0.1, the calculated value of y(1.2) equals 15.920, and for h=0.05, y(1.2) amounts to 16.523.

Step-by-step explanation:

Given,

with y(1)=9.

This means when x=1, y=9 initially.

To determine the value of y= f(x,y) at when h=0.1 and h=0.05, we will implement a C program using Euler's method resulting in, for h=0.1:

#include<stdio.h>

float fun(float x, float y)

{

  float f;

  f=2x+3y+1;

  return f;

}

main()

{

   float a,b,x,y,h,t,k;

   printf("\ Enter x0,y0,h,xn:");

   scanf("% f % f % f % f ", & a, &b, &h, &t);

   x=a;

   y=b;

   printf("\n x\t y\n");

   while (x<=t)

}

Using initial values x0=1, y0=9, h=0.1, and xn=1.2, we reach results, y(1.2)=15.920.

Subsequently, using values x0=1, y0=9, with h=0.05, and xn=1.2 yields y(1.2)=16.523.