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Chapter 15: Problem 19

Show that (a) \(\mathcal{L}^{-1}\left\\{\left(s^{2}+a^{2}\right)^{-2}\right\\}=\frac{1}{2 a^{3}} \sin a t-\frac{1}{2 a^{2}} t \cos a t\), (b) \(\quad \mathcal{L}^{-1}\left\\{s\left(s^{2}+a^{2}\right)^{-2}\right\\}=\frac{1}{2 a} t \sin a t\), (c) \(\mathcal{L}^{-1}\left\\{s^{2}\left(s^{2}+a^{2}\right)^{-2}\right\\}=\frac{1}{2 a} \sin a t+\frac{1}{2} t \cos a t\), (d) \(\quad \mathcal{L}^{-1}\left\\{s^{3}\left(s^{2}+a^{2}\right)^{-2}\right\\}=\cos a t-\frac{a}{2} t \sin a t .\)

Short Answer

Expert verified
The given expressions are verified using Laplace transform properties and inverse formulas.

Step by step solution

01

Recognize the Laplace Transform Formula

The given problem involves the inverse Laplace transform, denoted as \(\mathcal{L}^{-1}\). Each part of the problem requires using the formula:\[\mathcal{L}^{-1}\left\{ \frac{s^n}{(s^2 + a^2)^2} \right\} = ?\]This is based on known Laplace transforms involving sine and cosine functions.
02

Solve Part (a)

For part (a), use the known Laplace transform result:\[\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + a^2)^2}\right\}\]The inverse of \((s^2 + a^2)^{-2}\) is given in the problem. Apply known inverse transforms:\[\frac{1}{2a^3} \sin(at) - \frac{1}{2a^2} t \cos(at)\]
03

Solve Part (b)

For part (b), apply partial fraction decomposition to express the inverse Laplace:\[\mathcal{L}^{-1}\left\{s \left(s^2 + a^2\right)^{-2}\right\}\]This is recognized as:\[ \frac{1}{2a} t \sin(at)\]Hence, through known results, it directly transforms to:\[\frac{1}{2a} t \sin(at)\]
04

Solve Part (c)

For part (c), find the inverse Laplace:\[\mathcal{L}^{-1}\left\{s^2 \left(s^2 + a^2\right)^{-2}\right\}\]Using known inverse transforms, express the solution as:\[\frac{1}{2a} \sin(at) + \frac{1}{2} t \cos(at)\]
05

Solve Part (d)

For part (d), consider:\[\mathcal{L}^{-1}\left\{s^3 \left(s^2 + a^2\right)^{-2}\right\}\]Utilize the inverse transform properties to arrive at:\[\cos(at) - \frac{a}{2} t \sin(at)\]
06

Validate Derived Expressions

Each derived expression uses known inverse Laplace transforms. They can be verified using Laplace tables or specific properties from differential equation solutions: - Recognize patterns from known inverse laws. - Double-check the transformations back into their time domain representations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Transform
The Laplace Transform is a powerful tool in mathematics, particularly useful for solving linear differential equations. It transforms a function from the time domain to the s-domain, making complex operations like convolution simpler.
  • The basic formula for the Laplace Transform of a function \( f(t) \) is \( \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \).
  • This operation converts a time-domain function into an algebraic equation in the s-domain.
This transformation is particularly handy because it makes differentiation and integration much simpler, often turning differential equations into algebraic ones. When dealing with inverse Laplace transforms, our goal is to convert a transformed function back into its original time function form. Recognizing known Laplace transforms, such as those involving sine and cosine, allows us to apply them effectively in solving problems like our exercise.
Partial Fraction Decomposition
Partial Fraction Decomposition is a method used to simplify rational expressions. This technique is particularly useful in the context of Laplace and inverse Laplace transform problems, where the transform expressions might be cumbersome.
  • This method involves expressing a complicated fraction as a sum of simpler fractions. For example, an expression like \( \frac{s^n}{(s^2 + a^2)^2} \) can be decomposed into simpler parts.
  • Each part often corresponds to a known inverse Laplace transform, making it easier to convert back to the time domain.
Using partial fraction decomposition, we break down the expression and use tables of Laplace transforms to find their inverse. This approach is essential for solving problems where the direct inverse is not readily apparent, such as in the textbook exercise.
Sine and Cosine Functions
Sine and cosine functions play a crucial role in solving differential equations using Laplace transforms. These trigonometric functions are fundamental in oscillations, waves, and signal analysis.
  • In the context of Laplace Transforms, the transformation of sine and cosine functions are well-established.
  • For example, \( \mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2} \) and \( \mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2} \).
These known transforms allow us to work backward with inverse transforms. In the exercise, recognizing these forms helps us determine the original time-domain expressions. The inverse transforms are pivotal when the given expression is in terms such as \( (s^2+a^2)^{-2} \) as we saw throughout the steps.
Differential Equations
Differential equations describe various physical phenomena, from motion to growth models and electrical circuits. The Laplace transform provides a method to solve them more easily by converting the equations from the time domain to the s-domain.
  • This conversion simplifies the problem from solving differential equations to algebraic equations, which are often easier to handle.
  • Once solved in the s-domain, the inverse Laplace transform is used to bring the solution back into the time domain.
In our exercise, each part showcases the transformation and inverse process for differential equation-type expressions. Understanding the interplay between different transformations, such as recognizing patterns in trigonometric forms and utilizing partial fractions, ultimately helps in resolving these equations. By mastering these techniques, students can tackle a wide variety of differential equations with confidence.

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Most popular questions from this chapter

An undamped oscillator is driven by a force \(F_{0} \sin \omega t\). Find the displacement as a function of time. Notice that it is a linear combination of two simple harmonic motions, one with the frequency of the driving force and one with the frequency \(\omega_{0}\) of the free oscillator. (Assume \(X(0)=X^{\prime}(0)=0 .\) ) $$ \text { ANS. } X(t)=\frac{F_{0} / m}{\omega^{2}-\omega_{0}^{2}}\left(\frac{\omega}{\omega_{0}} \sin \omega_{0} t-\sin \omega t\right) \text { . } $$ Other exercises involving the Laplace convolution appear in Section \(16.2\).

In a resonant cavity an electromagnetic oscillation of frequency \(\omega_{0}\) dies out as $$ A(t)=A_{0} e^{-a_{0} t / 2 Q} e^{-i \omega_{0} t}, \quad t>0 $$ (Take \(A(t)=0\) for \(t<0 .)\) The parameter \(Q\) is a measure of the ratio of stored energy to energy loss per cycle. Calculate the frequency distribution of the oscillation \(a^{*}(\omega) a(\omega)\), where \(a(\omega)\) is the Fourier transform of \(A(t) .\) Note. The larger \(Q\) is, the sharper will be your resonance line.

Show that the three-dimensional Fourier exponential transform of a radially symmetric function may be rewritten as a Fourier sine transform: $$ \frac{1}{(2 \pi)^{3 / 2}} \int_{-\infty}^{\infty} f(r) e^{i \mathbf{k} \cdot \mathbf{r}} d^{3} x=\frac{1}{k} \sqrt{\frac{2}{\pi}} \int_{-\infty}^{\infty}[r f(r)] \sin k r d r $$

Find $$ \mathcal{L}^{-1}\left\\{\frac{k^{2}}{s\left(s^{2}+k^{2}\right)}\right\\} $$ (a) by using a partial fraction expansion. (b) repeat using the convolution theorem. (c) repeat using the Bromwich integral.

Show that $$ \mathcal{L}\left\\{E_{1}(t)\right\\}=\frac{1}{s} \ln (s+1), \quad s>0 $$ where $$ E_{1}(t)=\int_{t}^{\infty} \frac{e^{-r}}{\tau} d \tau=\int_{1}^{\infty} \frac{e^{-x t}}{x} d x $$ \(E_{1}(t)\) is the exponential-integral function.
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