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Q2E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 259

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

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.
$y'' (t) = \cos (t - y) + y^{2} (t); y (0) = 1, y' (0) = 0$

$x'_{1} (t) = x_{2} (t) x'_{2} (t) = \cos (t - x_{1}) + {x^{2}}_{1} (t) x_{1} (0) = 1 x_{2} (0) = 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: express the equation in form of x

Here given $y'' (t) = \cos (t - y) + y^{2} (t)$

Denote,

$x_{1} (t) = y (t) x_{2} (t) = y' (t)$

The equation transforms as;

$x'_{1} (t) = x_{2} (t) x'_{2} (t) = \cos (t - x_{1}) + {x^{2}}_{1} (t)$

Step 2: the initial conditions

The given initial conditions are $y (0) = 1, y' (0) = 0 .$

Initial conditions after transformations;

$x_{1} (0) = 1 x_{2} (0) = 0$

This is the required result.

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