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Use the result of Problem 28 to find the Laplace transform of the given function; \(a\) and \(b\) are real numbers and \(n\) is a positive integer. $$ t e^{a t} \sin b t $$

Short Answer

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01

Recall the Laplace Transform Formula

The Laplace Transform \(L[f(t)]\) of a function \(f(t)\) is defined as: $$ L[f(t)] = \int_0^{\infty} e^{-st} f(t) dt $$ where s is a complex variable.
02

State the Given Function

The given function is as follows: $$ f(t) = t e^{at} \sin bt $$ where \(a\) and \(b\) are real numbers, and t is a variable.
03

Use Problem 28 Result

Since Problem 28 should provide a technique or formula to find the Laplace transform of \(t e^{at} \sin bt\), we will need to apply that result here. Please provide the result from Problem 28 to proceed with the solution.
04

Compute the Laplace Transform of the Given Function

Using the result from Problem 28, we will calculate the Laplace transform of our given function \(f(t) = t e^{at} \sin bt\) to find \(L[f(t)]\). Please provide the result from Problem 28 to proceed with the solution.
05

Check and Simplify the Result

After applying the result of Problem 28 and computing the Laplace transform of our given function, we will check and simplify the result as needed. If any steps are unclear, please revise the explanation accordingly.

Key Concepts

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

Laplace Transform Formula
The Laplace Transform is an integral transform widely used to solve differential equations by converting them from the time domain into the frequency domain. It is particularly helpful in engineering and physics where systems are analyzed in terms of their responses to various inputs.

The formula for the Laplace Transform of a function f(t), where t represents time, is given by:

$$L[f(t)] = \int_0^{\infty} e^{-st} f(t) dt$$

Here, s is a complex variable, and the exponential function e^{-st} acts as a weight, exponentially decreasing the value of the function as t gets larger. This ensures that the integral converges for functions of exponential order. Understanding how to apply this formula is crucial when working with differential equations and in system analysis.
Complex Variables
Complex variables are used in many advanced mathematical concepts, including the Laplace Transform. In the context of the Laplace Transform, the complex variable is usually denoted by s and can be expressed as s = σ + iω, where σ is the real part, ω is the imaginary part, and i is the imaginary unit (i^2 = -1).

The use of complex variables extends the applicability of the Laplace Transform to a broader range of functions. By factoring in the exponential decay through e^{-st}, the transform helps to analyze systems that are subject to oscillations and damping, which are common in engineering fields such as control systems and signal processing.
Integral Transforms
Integral transforms are mathematical operations that convert a given function into another function via an integral. The purpose of these transforms is often to simplify a complicated problem, making it easier to solve. The Laplace Transform is one type of integral transform.

Other common integral transforms include the Fourier Transform, which is used to convert a time-domain signal into its frequency components, and the Mellin Transform, which is used in number theory and complex analysis. The choice of transform is dependent on the nature of the problem and the characteristics of the function being analyzed.

Integral transforms are powerful tools because they can harness properties in one domain (like convolution in the time domain becoming multiplication in the frequency domain) to make certain mathematical operations simpler to perform.
Exponential Functions
Exponential functions, typically of the form e^x, are essential in the study of the Laplace Transform. The letter e represents Euler's number, approximately equal to 2.71828, and it is the base of the natural logarithm. These functions are characterized by the fact that the rate of growth (or decay) is proportional to the value of the function at any point in time.

In the Laplace Transform formula, the exponential function e^{-st} plays a pivotal role by acting as a damping factor which permits the transformation of a wide array of functions, including those that are not otherwise integrable using traditional methods. The variable s in the exponential is a complex number, which allows the transform to account for oscillatory behavior, much like that seen in sinusoidal functions, when investigating the behavior of dynamic systems.

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Most popular questions from this chapter

Consider the initial value problem $$ y^{\prime \prime}+\gamma y^{\prime}+y=\delta(t-1), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ where \(\gamma\) is the damping coefficient (or resistance). (a) Let \(\gamma=\frac{1}{2} .\) Find the value of \(k\) for which the response has a peak value of \(2 ;\) call this value \(k_{1} .\) (b) Repeat part (a) for \(\gamma=\frac{1}{4}\). (c) Determine how \(k_{1}\) varies as \(\gamma\) decreases. What is the value of \(k_{1}\) when \(\gamma=0 ?\)

Recall that \(\cosh b t=\left(e^{b t}+e^{-b t}\right) / 2\) and \(\sinh b t=\left(e^{b t}-e^{-b t}\right) / 2 .\) In each of Problems 7 through 10 find the Laplace transform of the given function; \(a\) and \(b\) are real constants. $$ e^{a t} \cosh b t $$

Express the solution of the given initial value problem in terms of a convolution integral. \(y^{\mathrm{iv}}+5 y^{\prime \prime}+4 y=g(t) ; \quad y(0)=1, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=0\)

Let \(f\) satisfy \(f(t+T)=f(t)\) for all \(t \geq 0\) and for some fixed positive number \(T ; f\) is said to be periodic with period \(T\) on \(0 \leq t<\infty .\) Show that $$ \mathcal{L}\\{f(t)\\}=\frac{\int_{0}^{T} e^{-s t} f(t) d t}{1-e^{-s T}} $$

Use the Laplace transform to solve the given initial value problem. $$ y^{\mathrm{iv}}-4 y^{\prime \prime \prime}+6 y^{\prime \prime}-4 y^{\prime}+y=0 ; \quad y(0)=0, \quad y^{\prime}(0)=1, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=1 $$

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