Question

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. $$ t e^{a t} \cos b t $$

Short answer

Question: Find the Laplace transform of the function \(t e^{at} \cos bt\). Answer: The Laplace transform of the function \(t e^{at} \cos bt\) is \(L\{t e^{at} \cos bt\} = \frac{s^2 - 2as + a^2 + b^2}{(s^2 - 2as + a^2 + b^2)^2}\).

Step-by-step solution

Recall the definition of Laplace transform

Recall that the Laplace transform of a function f(t) is defined as: $$ L\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt $$ We will use this definition to find the Laplace transform of the given function \(t e^{at} \cos bt\).

Substitute the given function into the definition

Substitute \(f(t) = t e^{at} \cos bt\) into the definition of the Laplace transform: $$ L\{t e^{at} \cos bt\} = \int_{0}^{\infty} e^{-st} (t e^{at} \cos bt) dt $$

Combine the exponential terms

Combine the exponential terms \(e^{-st}\) and \(e^{at}\), resulting in a single exponential term: $$ L\{t e^{at} \cos bt\} = \int_{0}^{\infty} t e^{(a-s)t} \cos bt dt $$

Apply integration by parts

Apply integration by parts to solve the integral, let $$ u = t,~ dv = e^{(a-s)t} \cos bt dt $$ Then, we have $$ du = dt, ~v = \frac{e^{(a-s)t}(b \sin bt +(a-s)\cos bt)}{b^2 +(a-s)^2} $$ Now, apply integration by parts formula: $$ L\{t e^{at} \cos bt\} = \left[u v\right]_0^{\infty} - \int_{0}^{\infty} u dv $$

Evaluate the term at the limits

Evaluate the term \(uv\) at the limits and subtract the remaining integral: $$ L\{t e^{at} \cos bt\} = \lim_{t\to\infty} \frac{t e^{(a-s)t}(b \sin bt +(a-s)\cos bt)}{b^2 +(a-s)^2} - \frac{e^{(a-s)t}(b \sin bt +(a-s)\cos bt)}{b^2 +(a-s)^2}\Big|_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{(a-s)t}(b \sin bt +(a-s)\cos bt)}{b^2 +(a-s)^2} dt $$

Simplify the Laplace transform expression

Since the first term vanishes for the given conditions and the term evaluated at the lower limit is zero. We obtain $$ L\{t e^{at} \cos bt\} = - \int_{0}^{\infty} \frac{e^{(a-s)t}(b \sin bt +(a-s)\cos bt)}{b^2 +(a-s)^2} dt $$ Now, divide integral into two parts, then we have: $$ L\{t e^{at} \cos bt\} = - \int_{0}^{\infty} \frac{e^{(a-s)t}\cos bt}{b^2 +(a-s)^2}(a-s) dt - \int_{0}^{\infty} \frac{e^{(a-s)t}\sin bt}{b^2 +(a-s)^2}(b) dt $$

Recognize the Laplace transforms

Recognize that these two integrals are the Laplace transforms of \(\cos bt\) and \(\sin bt\). We have: $$ L\{t e^{at} \cos bt\} = -\frac{a-s}{(a-s)^2+b^2}L\{e^{(a-s)t} \cos{(bt)}\} - \frac{b}{(a-s)^2+b^2}L\{e^{(a-s)t} \sin{(bt)}\} $$ Using the Laplace transform for exponential multiplied with sine and cosine, we finally get: $$ L\{t e^{at} \cos bt\} = -\frac{a-s}{(a-s)^2+b^2}\frac{s-a}{(s-a)^2 + b^2} + \frac{b}{(a-s)^2+b^2}\frac{b}{(s-a)^2 + b^2} = \frac{-(a-s)(s-a)+(b)^2}{[(s-a)^2+b^2]^2} $$ Thus, the Laplace transform of the given function \(t e^{at} \cos bt\) is $$ L\{t e^{at} \cos bt\} = \frac{s^2 - 2as + a^2 + b^2}{(s^2 - 2as + a^2 + b^2)^2} $$

Key concepts

Differential Equations

Differential equations are mathematical equations that describe relationships involving rates of change and functions of one or more variables. They're a cornerstone of understanding phenomena in various scientific fields. These equations take into account the rates at which things change, which can include motion, growth, decay, and more. When we talk about the Laplace transform, it provides a method for solving linear differential equations more easily. By transforming a differential equation into an algebraic equation, we simplify the process to solve complex problems by working with simpler polynomial expressions. This technique is particularly useful for engineers and scientists who work with systems described by differential equations, like electrical circuits, mechanical systems, and population models.

Integration by Parts

The method of integration by parts is a powerful tool for solving integrals that involve the product of two functions. It is derived from the product rule for differentiation and provides a way to break down complex integrals into more manageable parts. The formula is given by \( \int u dv = uv - \int v du \). To effectively use this method, one must make a wise choice between what to set as \( u \) and \( dv \) so that the resulting integral is easier to compute. This technique is integral to finding the Laplace transforms of products of functions, such as \( t e^{at} \cos bt \) in the original exercise. By choosing \( u = t \) and \( dv = e^{(a-s)t} \cos bt dt \) we can break down the process into simpler parts, making the complex integration manageable.

Choosing u and dv:

• \( u = t \) – Easier to differentiate.
• \( dv = e^{(a-s)t} \cos bt dt \) – Leaves a function that is manageable to integrate.

Exponential Functions

Exponential functions are fundamental in the study of growth and decay processes. Represented generally as \( e^{kt} \), where \( k \) is a constant and \( t \) typically stands for time, these functions show how quantities grow or decay at a rate proportional to their current value. This characteristic makes them incredibly important when dealing with processes that have constant growth rates such as radioactive decay, population growth, and interest calculations. Within the context of Laplace transforms, exponential functions often appear and are manipulated to explore the behavior of systems over time. In our exercise, we see \( e^{at} \) signifying an exponential growth factor that, combined with a trigonometric function, models oscillating systems growing over time.

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) describe oscillatory motion and are used extensively in the analysis of waves, vibrations, and periodic phenomena. The function \( \cos bt \) in particular, depicts a wave with frequency \( b \) and, when combined with an exponential growth function, can model damped harmonic motion. In the case of Laplace transforms, these trigonometric functions are often found within the integral to be transformed, indicating an oscillatory component in the time domain. The transform will then yield a form that includes frequency components making it a vital technique in the study of signals and systems.