Problem 84 - Essential Calculus

Question

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=e^{a t} $$

Step-by-Step Solution

Answer

The Laplace Transform of the function \( f(t) = e^{at} \) is \( F(s) = \frac{1}{s-a} \) if \( s > a \) and exists nowhere else.

Step 1: Setting up the Integral

Using the definition of the Laplace transform, plug the function into the integral to obtain: \( F(s) = \int_0^\infty e^{-st}e^{at}dt \)

Step 2: Simplifying the Integral

By laws of exponents, simplify the integral to: \( F(s) = \int_0^\infty e^{(a-s)t}dt \)

Step 3: Computing the Integral

The integral can be solved with a straightforward application of the formula for integration of exponential functions: \( F(s) = \frac{1}{a-s}e^{(a-s)t} \Bigg|_0^\infty \) if \( s > a \) and exists nowhere else.

Step 4: Evaluating the Integral

Evaluate the integral to obtain: \( F(s) = \frac{1}{a-s} [ e^{(a-s)\cdot \infty} - e^{0} ] \) if \( s > a \) and exists nowhere else. This simplifies as \( F(s) = \frac{1}{s-a} \) if \( s > a \) and exists nowhere else considering that e^(x) where x is negative infinity becomes 0 and e^(0) is 1.

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They play a crucial role in various fields such as physics, engineering, and economics.

Their primary use is to describe changes. Think of weather patterns, population growth, or the motion of planets. In these situations, differentials help to model the change over time.

For simple understanding, consider an example of a differential equation:

• \( \frac{dy}{dt} = ky \)

In this case, \( y \) is the unknown function, and \( \frac{dy}{dt} \) is its derivative. Here, the constant \( k \) dictates the rate of change of \( y \). This particular form is common in exponential growth and decay problems.

Why are Laplace transforms tied to them? Laplace transforms help convert these time-domain differential equations into an algebraic form in the frequency domain, which is often easier to solve.

Improper Integral

An improper integral is an integral where the interval of integration is infinite or the function being integrated becomes infinite within the interval.

The Laplace transform is an example of improper integration because it involves integrating from 0 to infinity.

• \( \int_0^\infty f(t) dt \)

Here, the integration limit is infinite, indicating an improper integral.

The reason these integrals are useful is because they allow us to compute values that extend to infinity or involve singularities. The condition for convergence is crucial here. For example, the integral in the exercise converges only if \( s > a \). This ensures the exponent \( e^{(a-s)t} \) decays quickly enough to zero as \( t \) approaches infinity.

Understanding how limits affect integration is essential. Knowing whether an integral converges informs us if the solution to a problem is valid or not.

Exponential Functions

Exponential functions are expressions in which a variable appears in the exponent. They can be expressed as \( f(t) = e^{x} \). These functions are powerful for modeling rapid growth or decay processes.

In the Laplace transform example, being able to manipulate exponents is key. Consider the transformation of \( e^{-st}e^{at} \) which simplifies to \( e^{(a-s)t} \) due to the properties of exponents:

• Product Rule: \( a^m \cdot a^n = a^{m+n} \)

This simplification plays a pivotal role in solving integrals as seen in the Laplace transform.

Exponential functions have unique properties:

• Their rate of change is proportional to their current value.
• They grow or decay at rates determined by their exponent.

Understanding these properties helps in applying the Laplace transform to equations involving exponential functions.