Chapter 3: Q 3.6-13E (page 130)

Use the improved Euler’s method with tolerance to approximate the solution to $y' = 1 - y + y^{3}, y (0) = 0$ , at x= 1. For a tolerance of $ε = 0.003$ , use a stopping procedure based on the absolute error.

Short Answer

Expert verified

$ϕ (1) = 0.716984$

Step by step solution

Find the equation of approximation value.

Here, $y' = 1 - y + y^{3}, y (0) = 0$

For $ξ = 0.01$ , x=0, y₀ = 0 , $c = π$ , M = 10, h = 1

$F = f (x, y) = 1 - y + y^{3} G = f (x + h, y + hF) = 1 - (y + hF) + (y + hF)^{3} = 1 - \sin (y + h (1 - \sin y)) 1 - (y + h (1 - y + y^{3})) + (y + h (1 - y + y^{3}))^{3}$

Solve for x and y.

Apply initial points $x = 0, y = 0, h = 1$

$F (0, 0) = 1 G (0, 0) = 1$

$x = (x + h) y = x + \frac{h}{2} (F + G)$

x = 1

y = 1

Hence $ϕ (1) = y (1, 1) = 1$

Evaluate the value of x and y

$x = 0, y = 0, h = 0.5$

$F (0, 0) = 1 G (0, 0) = 0 . 625$

$x = 0 + 0.5 = 0.5 y = 0 + 0 . 25 (1 + 0 . 625) = 0.40625$

$ϕ (0.5) = 0.40625$

Determine the value of x and t for the conditions.

$x = 0.5, y = 0.40625, h = 0.5$

$F = 0.660798 G = 0.663095$

$x = 0.5 + 0.5 = 1 y = 0.40625 + 0 . 25 (0 . 660798 + 0 . 663095) = 0.737223$

$ϕ (1) = y (1, 0 . 5) = 0.737223$

Determine the all-other values.

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

$ϕ (1) = y (1, 0 . 25) = 0.719412 ϕ (1) = y (1, 0 . 125) = 0.716984$

Since,

$|y (1, 0 . 196350) - y (1, 0 . 392699)| = |1.09580 - 1.09229| = 0.00351 < 0.01$

$ϕ (1) = 0.716984$

Hence the solution is role="math" localid="1664310510661" $ϕ (1) = 0.716984$

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Use the improved Euler’s method with tolerance to approximate the solution to $y' = 1 - y + y^{3}, y (0) = 0$ , at x= 1. For a tolerance of $ε = 0.003$ , use a stopping procedure based on the absolute error.

Short Answer

Step by step solution

Find the equation of approximation value.

Solve for x and y.

Evaluate the value of x and y

Determine the value of x and t for the conditions.

Determine the all-other values.

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Use the improved Euler’s method with tolerance to approximate the solution to y'=1-y+y3,y(0)=0, at x= 1. For a tolerance ofε=0.003 , use a stopping procedure based on the absolute error.

Short Answer

Step by step solution

Find the equation of approximation value.

Solve for x and y.

Evaluate the value of x and y

Determine the value of x and t for the conditions.

Determine the all-other values.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Calculus

Mechanics Maths

Discrete Mathematics

Logic and Functions

Pure Maths

Study anywhere. Anytime. Across all devices.

Use the improved Euler’s method with tolerance to approximate the solution to $y' = 1 - y + y^{3}, y (0) = 0$ , at x= 1. For a tolerance of $ε = 0.003$ , use a stopping procedure based on the absolute error.