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. Sketch the graph of the solution. $y'' + y = 3 s i n 2 t - - 3 (s i n 2 t) u (t - 2 π); y (0) = 1, y' (0) = - 2$

On solving the given initial value problem using the method of Laplace transforms, the solution is $y (t) = \cos t - \sin 2 t - [2 \sin t - \sin 2 t] u (t - 2 π)$ and the corresponding graph is,

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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 domainto a function of a complex variables.

Step 2: Taking Laplace Transform of initial value Problem

Given initial value problem

$y'' + y = 3 \sin 2 t - 3 (\sin 2 t) u (t - 2 π)$

Where $y (0) = 1$ and $y' (0) = - 2$

Laplace Transform for the initial value problem

$L y'' (s) + L y (s) = L [3 \sin 2 t - 3 \sin 2 t u (t - 2 π)] s^{2} L y (s) - s L y (0) - y' (0) + L y (s) = \frac{3 \cdot 2}{s^{2} + 4} - \frac{3 \cdot 2 e^{- 2 π s}}{s^{2} + 4}$

$s^{2} L y (s) - s + 2 + L y (s) = \frac{6}{s^{2} + 4} - \frac{6 e^{- 2 π s}}{s^{2} + 4} (s^{2} + 1) L y (s) - (s - 2) = \frac{6}{s^{2} + 4} - \frac{6 e^{- 2 π s}}{s^{2} + 4}$

$L y (s) = \frac{s - 2}{s^{2} + 1} + \frac{6}{(s^{2} + 4) (s^{2} + 1)} - \frac{6 e^{- 2 π s}}{(s^{2} + 4) (s^{2} + 1)}$

Step 3: By Partial function method

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

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

Step 4: Laplace transform function

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

Hence, $y (t) = \cos t - \sin 2 t - [2 \sin t - \sin 2 t] u (t - 2 π)$

and the graph is

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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. Sketch the graph of the solution. $y'' + y = 3 s i n 2 t - - 3 (s i n 2 t) u (t - 2 π); y (0) = 1, y' (0) = - 2$

Step by Step Solution

Step 1: Definition

Step 2: Taking Laplace Transform of initial value Problem

Step 3: By Partial function method

Step 4: Laplace transform function

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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. Sketch the graph of the solution.y''+y=3sin2t--3(sin2t)u(t-2π);y(0)=1,y'(0)=-2

Step by Step Solution

Step 1: Definition

Step 2: Taking Laplace Transform of initial value Problem

Step 3: By Partial function method

Step 4: Laplace transform function

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solve the given initial value problem using the method of Laplace transforms. Sketch the graph of the solution. $y'' + y = 3 s i n 2 t - - 3 (s i n 2 t) u (t - 2 π); y (0) = 1, y' (0) = - 2$