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.
$y'' + 2 y' + 10 y = g (t); y (0) = - 1, y' (0) = 0, w h e r e g (t) = {\begin{cases} 10, 0 ⩽ t ⩽ 10, \\ 20, 10 < t < 20, \\ 0, 20 < t \end{cases}$

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

y (t) = - 2 e^{- t} \cos 3 t - \frac{2}{3} e^{- t} \sin 3 t + 1 - e^{- (t - 10)} \cos 3 (t - 10) - \frac{1}{3} e^{- (t - 10)} \sin 3 (t - 10) + u (t - 10) + 2 e^{- (t - 20)} \cos 3 (t - 20) + \frac{2}{3} e^{- (t - 20)} \sin 3 (t - 20) - 2 u (t - 20)

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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 equation into 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

$y'' + 2 y' + 10 y = g (t)$

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

$g (t) = 10, 0 ⩽ t ⩽ 10 = 20, 10 ⩽ t ⩽ 20$

Using rectangular and unit function we can write

$g (t) = 10 - 10 u (t - 10) + 20 u (t - 10) - 20 u (t - 20) = 10 + 10 u (t - 10) - 20 u (t - 20)$

Taking Laplace transform of initial value problem is

$L y'' (s) + 2 L y' (s) + 10 L y (s) = L [g (l)]$

$s^{2} L y (s) - s L y (0) - y' (0) + 2 s L y (s) - 2 y (0) + 10 y (s) = L ∣ 10 + 10 u (t - 10) - 20 u (t - 20)] s^{2} L y (s) + s - 0 + 2 s L y (s) + 2 + 10 L y (s) = \frac{10}{s} + \frac{10 c}{s} - \frac{20 e^{10 s}}{s} (s^{2} + 2 s + 10) L y (s) + (s + 2) = \frac{10}{s} + \frac{10 t^{- 11)}}{s} - \frac{20 e^{- 10 s}}{s}$

$L y (s) = - \frac{s + 2}{(s^{2} + 2 s + 10)} + \frac{10}{s (s^{2} + 2 s + 10)} + \frac{10 e^{- 10 s}}{s (s^{2} + 2 s + 10)} - \frac{20 e^{- 10 s}}{s (s^{2} + 2 s + 10)} = - \frac{s + 1}{{(s + 1)}^{2} + 9} - \frac{1}{{(s + 1)}^{2} + 9} + \frac{10}{s (s^{2} + 2 s + 10)} + \frac{10 e^{- 10 s}}{s (s^{2} + 2 s + 10)} - \frac{20 e^{- 10 s}}{s (s^{2} + 2 s + 10)}$

Using partial fraction

$\frac{10}{s (s^{2} + 2 s + 10)} = \frac{- s - 2}{s^{2} + 2 s + 10} + \frac{1}{s}$

Equation first becomes as

$L y (s) = - \frac{s + 1}{{(s + 1)}^{2} + 9} - \frac{1}{{(s + 1)}^{2} + 9} + \frac{- s - 2}{s^{2} + 2 s + 10} + \frac{1}{s} + [\frac{- s - 2}{s^{2} + 2 s + 10} + \frac{1}{s}] e^{- 10 s} - [\frac{- s - 2}{s^{2} + 2 s + 10} + \frac{1}{s}] 2 e^{- 20 s}$

Step 3: Take inverse Laplace transform we get

$y (t) = - e^{- t} \cos 3 t - \frac{1}{3} e^{- t} \sin 3 t - e^{- t} \cos 3 t - \frac{1}{3} e^{- t} \sin 3 t + 1 - e^{- (t - 10)} \cos 3 (t - 10) - \frac{1}{3} e^{- (t - 10)} \sin 3 (t - 10) + u (t - 10) + 2 e^{- (t - 20)} \cos 3 (t - 20) + \frac{2}{3} e^{- (t - 20)} \sin 3 (t - 20) - 2 u (t - 20) = - 2 e^{- t} \cos 3 t - \frac{2}{3} e^{- t} \sin 3 t + 1 - e^{- (t - 10)} \cos 3 (t - 10) - \frac{1}{3} e^{- (t - 10)} \sin 3 (t - 10) + u (t - 10) + 2 e^{- (t - 20)} \cos 3 (t - 20) + \frac{2}{3} e^{- (t - 20)} \sin 3 (t - 20) - 2 u (t - 20)$

hence

$y (t) = - 2 e^{- t} \cos 3 t - \frac{2}{3} e^{- t} \sin 3 t + 1 - e^{- (t - 10)} \cos 3 (t - 10) - \frac{1}{3} e^{- (t - 10)} \sin 3 (t - 10) + u (t - 10) + 2 e^{- (t - 20)} \cos 3 (t - 20) + \frac{2}{3} e^{- (t - 20)} \sin 3 (t - 20) - 2 u (t - 20)$

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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.
$y'' + 2 y' + 10 y = g (t); y (0) = - 1, y' (0) = 0, w h e r e g (t) = {\begin{cases} 10, 0 ⩽ t ⩽ 10, \\ 20, 10 < t < 20, \\ 0, 20 < t \end{cases}$

Step by Step Solution

Step 1: Define Laplace Transform

Step 2: Apply Laplace transform

Step 3: Take inverse Laplace transform we get

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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''+2y'+10y=g(t); y(0)=-1, y'(0)=0, where g(t)={10, 0⩽t⩽10,20 ,10<t<20,0, 20<t

Step by Step Solution

Step 1: Define Laplace Transform

Step 2: Apply Laplace transform

Step 3: Take inverse Laplace transform we get

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Pure Maths

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 2 y' + 10 y = g (t); y (0) = - 1, y' (0) = 0, w h e r e g (t) = {\begin{cases} 10, 0 ⩽ t ⩽ 10, \\ 20, 10 < t < 20, \\ 0, 20 < t \end{cases}$