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Q11E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 404

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

Use the convolution theorem to find the inverse Laplace transform of the given function. $\frac{s}{(s - 1) (s + 2)} [H i n t : \frac{s}{s - 1} = 1 + \frac{1}{s - 1}]$

The inverse Laplace transform for the given function by using the convolution theorem is. $y (t) = \frac{2 e^{- 2 t}}{3} + \frac{e^{t}}{3}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define convolution theorem

Let and be piecewise continuous on $[0, \infty)$ and of exponential orderand set, $F (s) = L {f} (s) and G (s) = L {g} (s)$ then,

$L {f * g} (s) = F (s) G (s),$ or

$L^{- 1} {F (s) G (s)} (t) = (f * g) (t)$

Step 2: Use the convolution theorem to obtain the inverse Laplace transform

Consider the given equation,

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

Let,

$\begin{matrix} y (s) = \frac{s}{(s - 1) (s + 2)} \\ = \frac{1}{s + 2} [1 + \frac{1}{s - 1}] \\ = \frac{1}{s + 2} + \frac{1}{(s - 1) (s + 2)} \end{matrix}$

Take inverse Laplace transform on both sides,

$L^{- 1} [y (s)] = L^{- 1} [\frac{1}{s + 2}] + L^{- 1} [\frac{1}{(s - 1) (s + 2)}]$ $\to (1)$

Hence, the convolution formula is, $L^{- 1} [f (s) \cdot g (s)] = f ⋆ g = \int_{0}^{t} f (t - v) g (v) d v$ , where

$f (s) = \frac{1}{s - 1}$ and $f (t) = e^{t}$

$g (s) = \frac{1}{s + 2}$ and $g (t) = e^{- 2 t}$

Thus, the equationcan be written as,

$\begin{matrix} y (t) = e^{- 2 t} + \int_{0}^{t} e^{t - v} \cdot e^{- 2 v} d v \\ = e^{- 2 t} + e^{t} \int_{0}^{t} e^{- 3 v} d v \\ = e^{- 2 t} + e^{t} {[\frac{e^{- 3 v}}{- 3}]}_{0}^{t} \\ = e^{- 2 t} - \frac{e^{t}}{3} [e^{- 3 t} - 1] \end{matrix}$

$\begin{matrix} y (t) = e^{- 2 t} - \frac{e^{- 2 t}}{3} + \frac{e^{t}}{3} \\ = \frac{2 e^{- 2 t}}{3} + \frac{e^{t}}{3} \end{matrix}$

Therefore, the inverse Laplace transform for the given function is. $y (t) = \frac{2 e^{- 2 t}}{3} + \frac{e^{t}}{3}$

Most popular questions for Math Textbooks

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

${(1 + e^{- t})}^{2}$ .

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.

$y'' + 5 y' + 6 y = g (t) : y (0) = 0, y' (0) = 2, w h e r e g (t) = {\begin{cases} 0, 0 ⩽ t < 1, \\ t, 1 < t < 5, \\ 1, 5 < t \end{cases}$

In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

$\begin{matrix} x^{'} {-y}^{'} =(sint)u(t-π); x(0)=1, \\ {x+y}^{'} =0; y(0)=1 \end{matrix}$

In Problems 11–20, determine the partial fraction expansion for the given rational function.

$\frac{- 8 s^{2} - 5 s + 9}{(s + 1) (s^{2} - 3 s + 2)}$

In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

$\begin{matrix} x^{''} =y+u(t-1); & x(0)=1, & x^{'} (0)=0 \\ y^{''} =x+1-u(t-1); & y(0)=0, & y^{'} (0)=0 \end{matrix}$

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