Question

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

Short answer

The Laplace transform of the given function \(e^{at} \sinh bt\) is: $$ L[e^{at} \sinh bt] = \frac{1}{2}\left(\frac{1}{s-(a+b)} - \frac{1}{s-(a-b)}\right) $$

Step-by-step solution

Rewrite the function using the relationship of \(\sinh bt\)

Replace \(\sinh bt\) in the given function using the relationship \(\sinh b t = \frac{e^{b t} - e^{-b t}}{2}\): $$ e^{at} \sinh bt = e^{at} \cdot \frac{e^{bt} - e^{-bt}}{2} $$

Simplify the function

Distribute \(e^{at}\) to both terms inside the parentheses and simplify: $$ e^{at} \cdot \frac{e^{bt} - e^{-bt}}{2} = \frac{1}{2}(e^{(a+b) t} - e^{(a-b) t}) $$

Compute the Laplace transform

Use the Laplace transform formula, \(L[f(t)] = \int_0^{\infty} e^{-s t} f(t) dt\), for each term inside the parentheses individually: For \(e^{(a+b) t}\) term: $$ L[e^{(a+b) t}] = \int_0^{\infty} e^{-s t} e^{(a+b) t} dt $$ For \(e^{(a-b) t}\) term: $$ L[e^{(a-b) t}] = \int_0^{\infty} e^{-s t} e^{(a-b) t} dt $$

Combine the terms and simplify the integrands

Combine the exponents in each integrand: For the \(e^{(a+b) t}\) term: $$ L[e^{(a+b) t}] = \int_0^{\infty} e^{-(s-(a+b)) t} dt $$ For the \(e^{(a-b) t}\) term: $$ L[e^{(a-b) t}] = \int_0^{\infty} e^{-(s-(a-b)) t} dt $$

Evaluate the integrals using integration by parts

Evaluate each integral using integration by parts (remembering to divide by 2, as in the simplified function): For the \(e^{(a+b) t}\) term: $$ L[e^{(a+b) t}] = \frac{1}{s-(a+b)} $$ For the \(e^{(a-b) t}\) term: $$ L[e^{(a-b) t}] = \frac{1}{s-(a-b)} $$

Combine the Laplace transform terms

Combine both Laplace transform terms and remember to divide by 2: $$ L\left[\frac{1}{2}(e^{(a+b) t} - e^{(a-b) t})\right] = \frac{1}{2}\left(\frac{1}{s-(a+b)} - \frac{1}{s-(a-b)}\right) $$ So, the Laplace transform of the given function \(e^{at}\sinh bt\) is: $$ L[e^{at} \sinh bt] = \frac{1}{2}\left(\frac{1}{s-(a+b)} - \frac{1}{s-(a-b)}\right) $$

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogous to the trigonometric functions, but they are related to hyperbolas instead of circles. They are often used when dealing with Laplace transforms and other mathematical transformations. The most common hyperbolic functions are hyperbolic sine and hyperbolic cosine, denoted as \(\sinh(x)\) and \(\cosh(x)\), respectively. These functions are defined using exponential functions:

• \(\sinh b t = \frac{e^{bt} - e^{-bt}}{2}\)
• \(\cosh b t = \frac{e^{bt} + e^{-bt}}{2}\)

Both have interesting properties and identities, similar to their trigonometric counterparts. For instance, \(\cosh^2(x) - \sinh^2(x) = 1\).

Hyperbolic functions are crucial in solving differential equations and in various fields of engineering and physics. They come into play naturally when you transform certain expressions using the Laplace transform. By manipulating them into exponential forms, as done in the original exercise, you can simplify computations significantly.

Exponential Functions

Exponential functions are written in the form \(e^{x}\), where \(e\) is the mathematical constant approximately equal to 2.71828. This type of function is key in many areas of mathematics and is particularly powerful when combined with Laplace transforms.

Exponential functions have unique properties:

• They are the only functions that are equal to their own derivatives.
• They grow rapidly and are used to model growth processes, like population increase or radioactive decay.

In the context of the original exercise, the hyperbolic and exponential functions combine through the transformation \(\sinh b t = \frac{e^{bt} - e^{-bt}}{2}\). This makes it possible to dissect complex expressions into more manageable parts before applying the Laplace transform.

Exponential functions also play a critical role in solving differential equations. They provide solutions in scenarios where quantities change at rates proportional to their current value.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It's used to integrate products of functions and is formalized with the formula:
\[ \int u \, dv = uv - \int v \, du \]
where \(u\) and \(dv\) are parts of the product you are integrating.

This method is particularly useful when handling functions that become simpler upon differentiation, such as polynomials multiplied by exponentials, logarithms, or trigonometric functions. In the original exercise, integration by parts helps with the evaluation of the Laplace transforms of each term.

To apply integration by parts effectively, choose \(u\) such that its derivative simplifies the problem, and select \(dv\) so it can be easily integrated. For terms like \(e^{(a+b)t}\), the exponential function's self-derivative property simplifies the integration significantly.

When executing the integral \(\int_0^{\infty} e^{-(s-(a"+(b")) t } dt\), because the Laplace transformation integrates with respect to \(t\) from 0 to \( \infty \), using integration by parts ensures the solution correctly accounts for each element involved.