Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem $y' = 1 - y, y (0) = 0$ , at x = 1. Compare these approximations to the actual solution $y = 1 + e^{- x}$ evaluated at x = 1.

$ϕ_{2} (1) = 0.6274 ϕ_{4} (1) = 0.6227$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the value of f2(x,y)

Here $y' = 1 - y, y (0) = 0$

Apply the chain rule.

$f_{2} (x, y) = \frac{\partial f}{\partial x} (x, y) + \frac{\partial f}{\partial y} (x, y) f (x, y)$

Since $f_{2} (x, y) = 1 - y$

$\frac{\partial f}{\partial x} (x, y) = 0 \frac{\partial f}{\partial y} (x, y) = - 1$

So, the equation is $f_{2} (x, y) = y - 1$

Step 2: Evaluate the values of f3(x,y) and f4(x,y)

Apply the same procedure as step 1

$f_{3} (x, y) = y - 1 f_{4} (x, y) = 1 - y$

Step 3: Apply the recursive formulas for order 2

The recursive formula is

$x_{n + 1} = x_{n} + h y_{n + 1} = y_{n} + h f (x_{n} + y_{n}) + \frac{h^{2}}{2!} f_{2} (x_{n} + y_{n}) + . . . . . \frac{h^{p}}{p!} f_{p} (x_{n} + y_{n}) x_{n + 1} = x_{n} + 0.25 y_{n + 1} = y_{n} + 0 . 25 (1 - y_{n}) + (\frac{{(0.25)}^{2}}{2}) (y_{n} - 1)$

Step 4: Apply the initial condition and find the values of x and y

Where starting points are $x_{o} = 0, y_{0} = 0$

$x_{1} = 0.25 y_{1} = 0.21875$

Put all these values in recursive formulas for the other values.

$x_{2} = 0.5 y_{2} = 0.389648 x_{3} = 0.75 y_{3} = 0.523163 x_{4} = 1 y_{4} = 0.627471$

Therefore the approximation of the solution by the Taylor method of order 2 at point x=1

$ϕ_{2} (1) = 0.6274$

Step 5: Apply the recursive formulas for order 4

$x_{n + 1} = x_{n} + 0.25 y_{n + 1} = y_{n} + 0 . 25 (1 - y_{n}) + (\frac{{(0.25)}^{2}}{2}) (y_{n} - 1) + \frac{{(0.25)}^{3}}{6} (y_{n} - 1) + \frac{{(0.25)}^{4}}{24} (1 - y_{n})$

Step 6: Apply the initial condition and find the values of x and y

Where starting points are $x_{o} = 0, y_{0} = 0$ .

$x_{1} = 0.25 y_{1} = 0.216309$

Put all these values in recursive formulas for the other values.

$x_{2} = 0.5 y_{2} = 0.385828 x_{3} = 0.75 y_{3} = 0.518679 x_{4} = 1 y_{4} = 0.622793$

Thus, the approximation of the solution by the Taylor method of order 4 at point x = 1

$ϕ_{4} (1) = 0.6227$

The actual solution at x = 1

$y (x) = 1 - e^{- x} y (1) = 0 . 632121$

Now combining the approximation with the actual solution at x = 1

$|y (1) - ϕ_{2} (1)| = 0.00465 |y (1) - ϕ_{4} (1)| = 0.00933$

Hence the solution is

$ϕ_{2} (1) = 0.6274 ϕ_{4} (1) = 0.6227$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem $y' = 1 - y, y (0) = 0$ , at x = 1. Compare these approximations to the actual solution $y = 1 + e^{- x}$ evaluated at x = 1.

Step by Step Solution

Step 1: Find the value of f2(x,y)

Step 2: Evaluate the values of f3(x,y) and f4(x,y)

Step 3: Apply the recursive formulas for order 2

Step 4: Apply the initial condition and find the values of x and y

Step 5: Apply the recursive formulas for order 4

Step 6: Apply the initial condition and find the values of x and y

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem y'=1-y,y(0)=0, at x = 1. Compare these approximations to the actual solution y=1+e-x evaluated at x = 1.

Step by Step Solution

Step 1: Find the value of f2(x,y)

Step 2: Evaluate the values of f3(x,y) and f4(x,y)

Step 3: Apply the recursive formulas for order 2

Step 4: Apply the initial condition and find the values of x and y

Step 5: Apply the recursive formulas for order 4

Step 6: Apply the initial condition and find the values of x and y

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Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem $y' = 1 - y, y (0) = 0$ , at x = 1. Compare these approximations to the actual solution $y = 1 + e^{- x}$ evaluated at x = 1.