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

solve the given initial value problem using the method of Laplace transforms.
$y'' + 2 y' + 2 y = u (t - 2 π) - - u (t - 4 π); y (0) = 1, y' (0) = 1$

On solving the given initial value problem using the method of Laplace transforms,the solution is

$y (t) = e^{- t} \cos t + 2 e^{- t} \sin t + \frac{1}{2} [1 - e^{- t + 2 π} [\cos t + \sin t]] u (t - 2 π) - \frac{1}{2} [1 - e^{- t + 4 π} [\cos t + \sin t]] u (t - 4 π)$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition

The Laplace transform, is an integral transform that converts a function of a real variable usually t, the time domain to a function of a complex variable s.

Step 2: Laplace transform function

Given initial value problem,

$y'' + 2 y' + 2 y = u (t - 2 π) - u (t - 4 π)$

Where $y (0) = 1$ and $y' (0) = 1$

Taking Laplace transform for the initial value problem

$L y'' (s) + 2 L y' (s) + 2 L y (s) = L [u (t - 2 π) - u (t - 4 π)] s^{2} L y (s) - s L y (0) - y' (0) + 2 s L y (s) - 2 y (0) + 2 L y (s) = \frac{e^{- 2 π s}}{s} - \frac{e^{- 4 π s}}{s}$

$s^{2} L y (s) - s - 1 + 2 s L y (s) - 2 + 2 L y (s) = \frac{e^{- 2 π s}}{s} - \frac{e^{- 4 π s}}{s} (s^{2} + 2 s + 2) L y (s) - (s + 3) = \frac{e^{- 2 π s}}{s} - \frac{e^{- 4 π s}}{s}$

$L y (s) = \frac{s + 3}{(s^{2} + 2 s + 2)} + \frac{e^{- 2 π s}}{s (s^{2} + 2 s + 2)} - \frac{e^{- 4 π s}}{s (s^{2} + 2 s + 2)} = \frac{s + 1 + 2}{{(s + 1)}^{2} + 1} + \frac{e^{- 2 π s}}{s (s^{2} + 2 s + 2)} - \frac{e^{- 4 π s}}{s (s^{2} + 2 s + 2)}$

Step 3: By partial function method

$\frac{1}{s (s^{2} + 2 s + 2)} = \frac{\frac{1}{2}}{s} - \frac{\frac{1}{2} s + 1}{s^{2} + 2 s + 2}$

$L y (s) = \frac{s + 1}{{(s + 1)}^{2} + 1} + \frac{2}{{(s + 1)}^{2} + 1} + \frac{\frac{e^{- 2 π s}}{2}}{s} - \frac{\frac{e^{- 2 π s}}{2} s + e^{- 2 π s}}{s^{2} + 2 s + 2} - \frac{\frac{e^{- 4 π s}}{2}}{s} + \frac{\frac{e^{- 4 π s}}{2} s + e^{- 4 π s}}{s^{2} + 2 s + 2}$

Step 4: Taking inverse Laplace transform

$y (t) = e^{- t} \cos t + 2 e^{- t} \sin t + \frac{1}{2} u (t - 2 π) - \frac{1}{2} e^{- t + 2 π} \cos (t - 2 π) u (t - 2 π) + \frac{1}{2} e^{- t + 2 π} \sin (t - 2 π) u (t - 2 π) - e^{- t + 2 π} \sin (t - 2 π) u (t - 2 π) - \frac{1}{2} u (t - 4 π) + \frac{1}{2} e^{- t + 4 π} \cos (t - 4 π) u (t - 4 π) - \frac{1}{2} e^{- t + 4 π} \sin (t - 4 π) u (t - 4 π) + e^{- t + 4 π} \sin (t - 4 π) u (t - 4 π)$

$y (t) = e^{- t} \cos t + 2 e^{- t} \sin t + \frac{1}{2} u (t - 2 π) - \frac{1}{2} e^{- t + 2 π} \cos t u (t - 2 π) - \frac{1}{2} e^{- t + 2 π} \sin t u (t - 2 π) - \frac{1}{2} u (t - 4 π) + \frac{1}{2} e^{- t + 4 π} \cos t u (t - 4 π) + \frac{1}{2} e^{- t + 4 π} \sin t u (t - 4 π) = e^{- t} \cos t + 2 e^{- t} \sin t + \frac{1}{2} [1 - e^{- t + 2 π} [\cos t + \sin t]] u (t - 2 π) - \frac{1}{2} [1 - e^{- t + 4 π} [\cos t + \sin t]] u (t - 4 π)$

hence

$y (t) = e^{- t} \cos t + 2 e^{- t} \sin t + \frac{1}{2} [1 - e^{- t + 2 π} [\cos t + \sin t]] u (t - 2 π) - \frac{1}{2} [1 - e^{- t + 4 π} [\cos t + \sin t]] u (t - 4 π)$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

solve the given initial value problem using the method of Laplace transforms.
$y'' + 2 y' + 2 y = u (t - 2 π) - - u (t - 4 π); y (0) = 1, y' (0) = 1$

Step by Step Solution

Step 1: Definition

Step 2: Laplace transform function

Step 3: By partial function method

Step 4: Taking inverse Laplace transform

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

solve the given initial value problem using the method of Laplace transforms.y''+2y'+2y=u(t-2π)--u(t-4π);y(0)=1,y'(0)=1

Step by Step Solution

Step 1: Definition

Step 2: Laplace transform function

Step 3: By partial function method

Step 4: Taking inverse Laplace transform

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solve the given initial value problem using the method of Laplace transforms.
$y'' + 2 y' + 2 y = u (t - 2 π) - - u (t - 4 π); y (0) = 1, y' (0) = 1$