Problem 16 - Differential Equations and Linear Algebra

Question

Use properties of the Laplace transform and the table of Laplace transforms to determine \(L[f]\). $$f(t)=2(t-5) u_{5}(t)$$

Step-by-Step Solution

Answer

The short answer: The Laplace transform of the given function \(f(t) = 2(t-5)u_5(t)\) is: \[L[f(t)] = \frac{2e^{-5s}}{s^2}\]

Step 1: Analyze the unit step function

The given function is written in terms of the unit step function \(u_5(t)\), which is defined as: $$u_5(t) = \begin{cases} 0, & \text{if}\ t < 5 \\ 1, & \text{if}\ t \ge 5 \end{cases}$$ It acts like a "switch" that turns on at \(t = 5\). So, in the given function \(f(t) = 2(t-5)u_5(t)\), the term \(u_5(t)\) ensures that the function is active only when \(t \ge 5\), i.e., the Laplace Transform will only consider the function values for \(t \ge 5\).

Step 2: Rewrite the function with the step function

Using the unit step function property, we can rewrite the given function as: $$f(t) = \begin{cases} 0, & \text{if}\ t < 5 \\ 2(t-5), & \text{if}\ t \ge 5 \end{cases}$$

Step 3: Recall the Laplace transform property for time delay

The Laplace transform of a delayed function follows the time-delay property: \[L[f(t-a)u_a(t)] = e^{-as}F(s) \Longrightarrow L[f(t)]\] where \(u_a(t)\) is the unit step function, and \(F(s)\) is the Laplace transform of the original function \(f(t)\) without delay.

Step 4: Apply the Laplace transform to the given function

Now, using the Laplace transform property for time delay, we can find the Laplace transform of the given function: \(L[f(t)] = L[2(t-5)u_5(t)]\) According to the time-delay property mentioned in step 3, we have: \(L[f(t)] = e^{-5s}L[2t]\) Now we just need to find the Laplace transform of the simpler function \(2t\).

Step 5: Find the Laplace transform of the simpler function

Using the Laplace transform table, we can find the Laplace transform of the function \(2t\): \(L[2t] = \frac{2}{s^2}\)

Step 6: Combine the results to find the Laplace transform of the original function

Now we can substitute the result from step 5 into our calculation from step 4: \(L[f(t)] = e^{-5s}L[2t] = e^{-5s}\frac{2}{s^2}\) So, the Laplace transform of the original function is: \[L[f(t)] = \frac{2e^{-5s}}{s^2}\]

Key Concepts

Unit Step Function

The unit step function plays a critical role in understanding piecewise functions in mathematics. Practically, you can think of it as a switch that changes the state at a specific time.
In mathematical operations, this function is often denoted by \( u_a(t) \), where "a" is the time at which the function transitions from 0 to 1. For example, \( u_5(t) \) remains 0 until \( t = 5 \) and becomes 1 afterwards.

• If \( t < 5 \), \( u_5(t) = 0 \)
• If \( t \geq 5 \), \( u_5(t) = 1 \)

This function is particularly useful in Laplace transforms because it defines when a certain function or condition should begin impacting the result. In problem-solving, recognizing this step is crucial because it can be used to simplify the treatment of function inputs over different intervals.

Time-Delay Property

A valuable property in the Laplace transform is the time-delay property. This property allows us to handle functions that activate after a certain time.
The time-delay property is expressed as: \[ L[f(t-a)u_a(t)] = e^{-as}F(s) \]Here, \( f(t) \) is your original function and \( u_a(t) \) acts as a toggle which starts the function at \( t = a \).
When you apply the Laplace transform to the delayed and scaled function, the transform is scaled by an exponential term, \( e^{-as} \), where "a" corresponds to the delay.

• The delay, \( a \), essentially shifts the start of the signal.
• Scaling by \( e^{-as} \) adjusts the response accordingly in the Laplace domain.

This property is immensely useful as it allows for simple mathematical manipulation instead of manually integrating over the active time interval. In the presented problem, we used this property to manage the time delay expressed by \( u_5(t) \), having a shift "a" equal to 5.

Laplace Transform Table

The Laplace transform table is a critical tool for quickly finding transforms of standard functions without calculating integrals every time.
This table lists Laplace transforms for basic functions like sinusoids, exponentials, and polynomials. Using these, we can solve more complex problems by breaking them down into simpler components.

• The table helps match the form of a function, allowing substitution of known transforms into our calculations.
• For instance, the transform of \( 2t \) is extracted as \( \frac{2}{s^2} \) from this table, allowing an effective solution for the example given.

In practice, having familiarity with these standard entries accelerates problem-solving. In this problem, leveraging the Laplace transform table simplified the otherwise complex process of finding the transform of the given function.