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Q 3.7-3E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 139

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

Determine the recursive formulas for the Taylor method of order 4 for the initial value problem $y' = x - y, y (0) = 0$ .

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

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 $y' = x - y, y (0) = 0$

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 $f (x, y) = x - y$

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

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

Step 2: Evaluate the values of f2(x,y) and f4(x,y)

Apply the same procedure as step 1

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

Step 3: Apply the recursive formulas for order 4

The recursive formula is

$x_{n + 1} = x_{n} + h y_{n + 1} = y_{n} + hf (x_{n} + y_{n}) + \frac{h^{2}}{2!} f_{2} (x_{n} + y_{n}) + . . . . . \frac{h^{p}}{p!} f_{p} (x_{n} + y_{n})$

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

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

Hence the solution is role="math" localid="1664316175651" $y_{n + 1} = y_{n} + h (x_{n} - y_{n}) + (\frac{h^{2}}{2} - \frac{h^{3}}{6} + \frac{h^{4}}{24}) (1 - x_{n} + y_{n})$

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