Q 3.6-12E

Question

Use the improved Euler’s method with tolerance to approximate the solution to y'=1-siny,y(0)=0, at  x=π. For a tolerance of  ε=0.01, use a stopping procedure based on the absolute error.

Step-by-Step Solution

Verified
Answer

The required result is  ϕπ=1.09580

1Step 1: Important formula.

The required Euler’s formula,

  x=x+hy=x+h2F+G

2Step 2: Find the equation of approximation value

Here given y'=1-siny,y0=0 ,


For value of ε=0.01 , x = 0, y0 = 0, c=π , M = 10, h = 3.141593 then

F=fx,y=1-sinyG=fx+h,y+hF=1-siny+hF=1-siny+h1-siny


3Step 3: solve for x and y

Apply initial points  

 F0,0=1G0,0=1

x=x+hy=x+h2F+Gx=3.141593y=3.14159

Hence, the value is ϕπ=y1,3.141593=3.14159

 

 

4Step 4: Evaluate the value of x and y

Now, for the values of x=0,y=0,h=1.570796

F0,0=1G0,0=5.34017×10-14

dx=0+1.570796=1.570796y=0+1.57079621+5.34017×10-14=0.785398ϕ1.570796=0.785398



5Step 5: Determine the value of x and t for the conditions

Now, for the values of F and G


x=1.570796,y=0.785398,h=1.570796


F1.570796,0.785398=0.292893G1.570796,0.785398=0.0524523x=1.570796+0.570796=3.14159y=1.05663

The value of  ϕπ=y1,0.570796=1.05663

6Step 6: Determine the all-other values.

Apply the same procedure for all other values and the values are 

ϕπ=y1,0.785398=1.07575ϕ(π)=y1,0.392699=1.09229ϕ(π)=y1,0.196350=1.09580

Since the value is 

y1,0.196350-y1,0.392699=1.09580-1.09229=0.00351<0.01ϕπ=1.09580




Therefore, the result is  ϕπ=1.09580