Problem 14 - Differential Equations and Linear Algebra

Question

Use the linearity of \(L\) and the formulas derived in this section to determine \(L[f]\). $$f(t)=2 t-e^{2 t}$$

Step-by-Step Solution

Answer

The Laplace Transform of the function \(f(t) = 2t - e^{2t}\) is: \[L[f(t)] = \frac{2}{s^2} - \frac{1}{s - 2}\]

Step 1: Apply the Linearity Property to \(f(t)\)

The linearity property of the Laplace Transform states that if you have a linear combination of functions, the Laplace Transform of this combination is equal to the same linear combination of the individual Laplace Transforms. In mathematical terms, for any functions \(f(t)\) and \(g(t)\) and any constants \(a\) and \(b\), we have: \[L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]\] Applying this property to the given function \(f(t) = 2t - e^{2t}\), we get: \[L[f(t)] = L[2t - e^{2t}] = 2L[t] - L[e^{2t}]\]

Step 2: Find the Laplace Transform of the first term (\(2t\))

Using the standard Laplace Transform formula for a function of the form \(t^n\), where \(n = 1\), we have: \[L[t] = \frac{n!}{s^{n+1}} = \frac{1!}{s^{1+1}} = \frac{1}{s^2}\]

Step 3: Find the Laplace Transform of the second term (\(-e^{2t}\))

Using the standard Laplace Transform formula for an exponential function of the form \(e^{at}\), where \(a = 2\), we have: \[L[e^{2t}] = \frac{1}{s - a} = \frac{1}{s - 2}\]

Step 4: Combine the results using the linearity property

We found that: \[L[f(t)] = 2L[t] - L[e^{2t}] = 2\left(\frac{1}{s^2}\right) - \left(\frac{1}{s - 2}\right)\]

Step 5: Write the final result

The Laplace Transform of the function \(f(t) = 2t - e^{2t}\) is: \[L[f(t)] = \frac{2}{s^2} - \frac{1}{s - 2}\]

Key Concepts

Linearity Property

The Linearity Property is fundamental in understanding how the Laplace Transform handles linear combinations of functions. It simplifies complex transforms by allowing you to deal with each component separately. Imagine you have two functions, say \(f(t)\) and \(g(t)\), combined linearly as \(af(t) + bg(t)\). This property tells us that you can transform each function separately and then combine them back using the same coefficients \(a\) and \(b\).
Here’s what this looks like mathematically:

• If \(L[f(t)]\) is the Laplace of \(f(t)\) and \(L[g(t)]\) is the Laplace of \(g(t)\), then \(L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]\).

This property is incredibly useful in breaking down complex functions into simpler parts, each of which can be easily transformed, and then combined back into a single result. For example, in our exercise, the function \(f(t) = 2t - e^{2t}\) was split, allowing us to handle \(2t\) and \(-e^{2t}\) separately, making the process far more straightforward.

Exponential Function

Exponential functions play a crucial role in the Laplace Transform due to their simplicity and frequent occurrence in differential equations. An exponential, in mathematics, is expressed as \(e^{at}\), where \(a\) is a constant, making it one of the simplest forms to convert using Laplace Transforms.
The formula for the Laplace Transform of an exponential function \(e^{at}\) is:

• \(L[e^{at}] = \frac{1}{s - a}\)

Where \(s\) is a complex frequency parameter often used in Laplace Transforms. This straightforward formula allows for easy transformation and is essential when solving differential equations involving exponential growth or decay. For instance, in our original problem, \(e^{2t}\) transforms to \(\frac{1}{s - 2}\), illustrating how the exponential term impacts the final result. Understanding this transformation helps more clearly analyze systems with exponential behaviors.

Standard Laplace Transform Formula

The Standard Laplace Transform Formula provides a set of rules or "cheat sheet" formulas that make converting functions from the time domain to the s-domain straightforward. These formulas are critical tools because, rather than recalculating transforms from scratch, you can refer to these pre-defined conversions.
Some basic formulas include:

• For constant \(C: L[C] = \frac{C}{s}\)
• For \(t^n: L[t^n] = \frac{n!}{s^{n+1}}\) (used when transforming polynomials)
• For \(e^{at}: L[e^{at}] = \frac{1}{s-a}\)

These formulas directly apply to frequently encountered situations, saving time and reducing error potential. In our exercise, using the polynomial formula \(L[t^n] = \frac{n!}{s^{n+1}}\), where \(n=1\) for \(t\), provided the transform \(\frac{1}{s^2}\).
This systematic and formulaic approach to Laplace Transforms assists in seamlessly managing each component of a function, ensuring efficiency and clarity in solving mathematical problems.