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 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = t u (t - 2) y (0) = 0 y' (0) = 1$

The solution of the given initial value problem using the method of Laplace transforms is.

$y (t) = e^{- 2 t} - e^{- 3 t} + [\frac{7}{36} + \frac{t - 2}{6} - \frac{3}{4} e^{- 2 (t - 2)} + \frac{5}{9} e^{- 3 (t - 2)}] u (t - 2)$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equations into algebraic equations. The formula for Laplace transform is

$F (s) = \int_{0}^{+ \infty} f (t) \cdot e^{- s \cdot t} d t$

Where, F(s) = Laplace Transform

S= complex number

t= real number >=0

t'= first derivative of the function f(t)

Step 2: Apply Laplace transform

Given initial value problem

$y'' + 5 y' + 6 y = t u (t - 2)$

where. $y (0) = 0 a n d y' (0) = 1$

Taking Laplace transform of initial value problem is

$L y'' (s) + 5 L y' (s) + 6 L y (s) = L [t u (t - 2)]$

$s^{2} L y (s) - s L y (0) - y' (0) + 5 s L y (s) - 5 y (0) + 6 L y (s) = e^{- 2 s} [\frac{1}{s^{2}} + \frac{2}{s}] s^{2} L y (s) - 0 - 1 + 5 s L y (s) - 0 + 6 L y (s) = e^{- 2 s} [\frac{1 + 2 s}{s^{2}}] (s^{2} + 5 s + 6) L y (s) - 1 = e^{- 2 s} [\frac{1 + 2 s}{s^{2}}]$

$L y (s) = \frac{1}{(s^{2} + 5 s + 6)} + e^{- 2 s} [\frac{1 + 2 s}{s^{2} (s^{2} + 5 s + 6)}] = \frac{1}{s + 2} - \frac{1}{s + 3} + e^{- 2 s} [\frac{1 + 2 s}{s^{2} (s^{2} + 5 s + 6)}] \dots (1)$

Using partial fraction

$\frac{1 + 2 s}{s^{2} (s^{2} + 5 s + 6)} = \frac{\frac{7}{36}}{s} + \frac{\frac{1}{6}}{s^{2}} - \frac{\frac{3}{4}}{s + 2} + \frac{\frac{5}{9}}{(s + 3)}$

Equation first becomes as

$L y (s) = \frac{1}{s + 2} - \frac{1}{s + 3} + \frac{\frac{7}{36} e^{- 2 s}}{s} + \frac{\frac{1}{6} e^{- 2 s}}{s^{2}} - \frac{\frac{3}{4} e^{- 2 s}}{s + 2} + \frac{\frac{5}{9} e^{- 2 s}}{(s + 3)}$

Taking inverse Laplace transform we get

$y (t) = e^{- 2 t} - e^{- 3 t} + \frac{7}{36} u (t - 2) + \frac{t - 2}{6} u (t - 2) - \frac{3}{4} e^{- 2 (t - 2)} u (t - 2) + \frac{5}{9} e^{- 3 (t - 2)} u (t - 2)$

Hence,

$y (t) = e^{- 2 t} - e^{- 3 t} + [\frac{7}{36} + \frac{t - 2}{6} - \frac{3}{4} e^{- 2 (t - 2)} + \frac{5}{9} e^{- 3 (t - 2)}] u (t - 2)$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = t u (t - 2) y (0) = 0 y' (0) = 1$

Step by Step Solution

Step 1: Define Laplace Transform

Step 2: Apply Laplace transform

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.y''+5y'+6y=tu(t-2)y(0)=0 y'(0)=1

Step by Step Solution

Step 1: Define Laplace Transform

Step 2: Apply Laplace transform

Most popular questions for Math Textbooks

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In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = t u (t - 2) y (0) = 0 y' (0) = 1$