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)

At h=0.1 value od y(1.2)=15.920, and at h=0.05, y(1.2)=16.523.

Step-by-step explanation:

Given,

with y(1)=9.

That is when x=1 then y=9 initially.

To find value of y= f(x,y) at when y=0.1 and y=0.05 we will use c-programme  of Eular method we get, 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);

}

}

By putting value x0=1, y0=9, h=0.1, xn=1.2  we will get results as, y(1.2)=15.920.

Again by putting values x0=1, y0=9, 0.h=05, xn=1.2 we get y(1.2)=16.523.

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.