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

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

yn + 1 = yn + hf(xn, yn) (3) by hand, first using h = 0.1 and then using h = 0.05. y' = 2x − 3y + 1, y(1) = 9; y(1.2) y(1.2) ≈ (h = 0.1) y(1.2) ≈ (h = 0.05)

Mathematics

1 answer:

AnnZ [9.1K] · Answer 1 · 2026-03-22T14:59:57+03:00

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,

$f(x,y)=\frac{dy}{dx}=2x+3y+1$ with y(1)=9.

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

To determine the value of y= f(x,y) at $x_n=x_0+nh=1.2$ 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)

{

k=h*fun(x,y);

x=x+h;

y=y+k;

printf("%0.3f\%0.3f\n",x,y);

}

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.