Chapter 12: Problem 24
a) Given that \(F(s)=\mathscr{L}\\{f(t)\\},\) show that \\[-\frac{d F(s)}{d s}=\mathscr{L}\\{t f(t)\\}\\] b) Show that \\[(-1)^{n} \frac{d^{n} F(s)}{d s^{n}}=\mathscr{L}\left\\{t^{n} f(t)\right\\}\\] c) Use the result of (b) to find \(\mathscr{L}\left\\{t^{5}\right\\}, \mathscr{L}\\{t \sin \beta t\\}\) and \(\mathscr{L}\left\\{t e^{-t} \cosh t\right\\}\)
Short Answer
Step by step solution
Understanding the Laplace Transform Property
Applying Integration by Parts to Laplace Transform
Generalization for Higher Powers
Application of (b) for \(\mathscr{L}\{t^5\}\)
Application of (b) for \(\mathscr{L}\{t \sin \beta t\}\)
Application of (b) for \(\mathscr{L}\{t e^{-t} \, \cosh t\}\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation Property
This property greatly simplifies solving problems where functions are multiplied by \( t \), as it avoids the direct computation of the Laplace transform from scratch.
To derive this result, integration by parts is crucial, allowing conversion of the time-domain multiplication by \( t \) into the frequency-domain operation of differentiation in \( s \). This highlights the powerful interplay between differentiation in the frequency domain and multiplication in the time domain in Laplace Transforms.
Integration by Parts
The rule states that for functions \( u(t) \) and \( v(t) \), the integral \( \int u \, dv = uv - \int v \, du \) can break down complex integrals into more manageable components.
In the context of Laplace transforms, one might let \( u = t \) and \( dv = f(t)e^{-st}dt \), resulting in substitutions \( du = dt \) and \( v = -e^{-st}/s \), which lead to the differentiation property mentioned earlier. This transformation simplifies evaluating functions in terms of their Laplace Transform properties.
Higher Powers of Laplace Transform
For example, to find the Laplace Transform of \( t^5 \), apply the above expression by differentiating five times and multiplying by \( (-1)^5 \). This results in computing \( \frac{5!}{s^6} \) for \( \mathscr{L}\{t^5\} \).
This generalization allows tackling transformations with higher powers easily by taking successive derivatives, transforming the complexity of time-domain power multiplication into repetitive frequency-domain differentiation.
Solve Laplace Transform Examples
- Example 1: Finding \( \mathscr{L}\{t \sin \beta t\} \).
Start with the known transform of \( \sin \beta t \), which is \( \frac{\beta}{s^2 + \beta^2} \). Differentiating this yields the transform of \( t \sin \beta t \), resulting in \( \frac{2s\beta}{(s^2 + \beta^2)^2} \). - Example 2: Compute \( \mathscr{L}\{t e^{-t} \cosh t\} \).
First, find \( F(s) \) by considering the Laplace transform of \( e^{-t} \cosh t \). Subsequent differentiation gives \( \mathscr{L}\{t e^{-t} \cosh t\} \), highlighting the application of expressed rules and operations.