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 3-10, determine the Laplace transform of the given function. $f (t) = c o s t, - π / 2 \leq t \leq π / 2$ and $f (t)$ has period $π$

Therefore, the solution is. $L {f (t)} = \frac{s}{s^{2} + 1} + \frac{2 e^{- \frac{π}{2} s}}{(s^{2} + 1) (1 - e^{- π s})}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

The given function value is. $f (t) = c o s t, - π / 2 \leq t \leq π / 2$

Step 2: Find the Laplace transform

On the interval we can write the given function as

$f (t) = | \cos t | = {\begin{cases} \cos t, & 0 \leq t \leq \frac{π}{2} \\ - \cos t, & \frac{π}{2} \leq t \leq π \end{cases}$

We will find the Laplace transform of given function using the following equation

$L {f (t)} (s) = \frac{\int_{0}^{T} e^{- s t} f (t) d t}{1 - e^{- s T}}$

Since . $T = π$ So, plug $T = π$ into above equation as:

$L {f (t)} (s) = \frac{\int_{0}^{π} e^{- s t} f (t) d t}{1 - e^{- s π}}$

Let's find $I = \int_{0}^{π} e^{- s t} f (t) d t .$

$\begin{matrix} I = \int_{0}^{π} e^{- s t} f (t) d t \\ = \underset{I_{1}}{\underset{⏟}{\int_{0}^{\frac{π}{2}} e^{- s t} f (t) d t}} - \underset{I_{2}}{\underset{⏟}{\int_{\frac{π}{2}}^{π} e^{- s t} f (t) d t}} \end{matrix}$ $\dots . (1)$

Step 3:Find I1and I2

The expression for $I_{1}$ is obtained as follows:

$\begin{matrix} I_{1} = \int_{0}^{\frac{π}{2}} e^{- s t} f (t) d t \\ = \int_{0}^{\frac{π}{2}} e^{- s t} \cos t d t \\ = | \begin{matrix} u = \cos t & d v = e^{- s t} d t \\ d u = - \sin t d t & v = - \frac{e^{- s t}}{s} \end{matrix} | \\ = {[- \frac{e^{- s t} \cos t}{s}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{e^{- s t} \sin t}{s} d t \end{matrix}$

Simplify further as:

$\begin{matrix} I_{1} = \frac{1}{s} - \frac{1}{s} \int_{0}^{\frac{π}{2}} e^{-st} sintdt \\ = | \begin{matrix} u=sint & {dv=e}^{-st} dt \\ du=costdt & v=- \frac{e^{-st}}{s} \end{matrix} | \\ = \frac{1}{s} - \frac{1}{s} ({[- \frac{e^{-st} sint}{s}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \frac{e^{-st} cost}{s} dt) \\ = \frac{1}{s} - \frac{1}{s} (- \frac{e^{- \frac{π}{2} t}}{s} + \frac{1}{s} I_{1}) \\ = \frac{1}{s} + \frac{e^{- \frac{π}{2} t}}{s^{2}} - \frac{1}{s^{2}} I_{1} \end{matrix}$

Step 4: Simplify the resulting equation

From the resulting equation that

$I_{1} = \frac{s + e^{- \frac{π}{2} s}}{s^{2} + 1}$ …… (2)

Similarly, We find

$I_{2} = \frac{e^{- \frac{π}{2} s} - e^{- π s} s}{s^{2} + 1}$ …… (3)

On substituting equation (2) and (3) in equation (1) gives

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

Thus the Laplace transform of given function is obtained as

$L {f(t)} = \frac{s ({1-e}^{-πs}) {+2e}^{- \frac{π}{2} s}}{(s^{2} +1) ({1-e}^{-πs})}$

Rewrite above equation as:

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

In Problems 3-10, determine the Laplace transform of the given function. $f (t) = c o s t, - π / 2 \leq t \leq π / 2$ and $f (t)$ has period $π$

Step by Step Solution

Step 1: Given Information

Step 2: Find the Laplace transform

Step 3:Find I1and I2

Step 4: Simplify the resulting equation

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 3-10, determine the Laplace transform of the given function.f(t)=cost,-π/2≤t≤π/2and f(t)has period π

Step by Step Solution

Step 1: Given Information

Step 2: Find the Laplace transform

Step 3:Find I1and I2

Step 4: Simplify the resulting equation

Most popular questions for Math Textbooks

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In Problems 3-10, determine the Laplace transform of the given function. $f (t) = c o s t, - π / 2 \leq t \leq π / 2$ and $f (t)$ has period $π$