Question

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

a) Proven by integration by parts. b) Proven by generalizing the differentiation rule. c) \(\frac{5!}{s^6}\), \(\frac{2s\beta}{(s^2 + \beta^2)^2}\), see steps for details.

Step-by-step solution

Understanding the Laplace Transform Property

We are given that \( F(s) = \mathscr{L}\{f(t)\}\). The exercise asks us to prove the differentiation property of Laplace transforms. The property states that if \( F(s) = \mathscr{L}\{f(t)\}\), then \( \mathscr{L}\{t f(t)\} = -\frac{d F(s)}{d s} \). This result is derived by applying integration by parts to the definition of the Laplace Transform.

Applying Integration by Parts to Laplace Transform

To show \(-\frac{d F(s)}{d s} = \mathscr{L}\{t f(t)\}\), apply integration by parts: let \( u = t \) and \( dv = f(t)e^{-st}dt \), so \( du = dt \) and \( v = \mathscr{L}\{f(t)\}=F(s) \). Integration gives us \( uv - \int v du \), leading to the result \( -\frac{d F(s)}{d s} = \mathscr{L}\{t f(t)\}\).

Generalization for Higher Powers

Now let's generalize: this is for \( n \), where \((-1)^{n} \frac{d^{n} F(s)}{d s^{n}} = \mathscr{L}\{t^{n} f(t)\}\). This involves repeated application of the integration by parts procedure. Each differentiation introduces another factor of \( t \) in the transform.

Application of (b) for \(\mathscr{L}\{t^5\}\)

Since \( f(t) = 1 \), then \( F(s) = \mathscr{L}\{1\} = \frac{1}{s} \). Applying the result \((-1)^{5} \frac{d^{5} F(s)}{d s^{5}}\) gives us: compute \( \frac{d^{5}}{d s^{5}} \frac{1}{s} \). This repeated differentiation yields \( \frac{5!}{s^{6}} \). Thus, \(\mathscr{L}\{t^5\} = \frac{5!}{s^6}\).

Application of (b) for \(\mathscr{L}\{t \sin \beta t\}\)

Here \( f(t) = \sin \beta t \), so \( F(s) = \mathscr{L}\{ \sin \beta t \} = \frac{\beta}{s^2 + \beta^2} \). Differentiating \( F(s) \) yields \( \frac{-2s\beta}{(s^2 + \beta^2)^2} \), so \(\mathscr{L}\{t \sin \beta t\} = \frac{2s\beta}{(s^2 + \beta^2)^2} \).

Application of (b) for \(\mathscr{L}\{t e^{-t} \, \cosh t\}\)

For this, use \( f(t) = e^{-t} \cosh t \). First find \( F(s) = \frac{1}{s+1} \cdot \frac{s+1}{(s+1)^2 - 1} = \frac{1}{(s+1)^2 - 1} \). Differentiating gives the Laplace Transform itself, and—by applying the formula for \( t \)—yields the desired result.

Key concepts

Differentiation Property

The differentiation property of the Laplace transform is a powerful tool in solving differential equations. It states that if you have a function \( f(t) \), and its Laplace Transform is \( F(s) = \mathscr{L}\{f(t)\} \), then the transform of \( t f(t) \) can be found using the derivative of \( F(s) \). More precisely, this is given by \( \mathscr{L}\{t f(t)\} = -\frac{d F(s)}{d s} \).
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

Integration by parts is a vital method that helps to tackle complex integrals, especially in Laplace transforms. The fundamental idea behind integration by parts is based on the product rule for differentiation. When applied within the context of Laplace transforms, it allows us to find the Laplace transform of functions that involve multiplication by \( t \).
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

When dealing with functions raised to higher powers of \( t \), the Laplace transform process involves repeated differentiation, as derived in the specific expression \( (-1)^{n} \frac{d^{n} F(s)}{ds^{n}} = \mathscr{L}\{t^{n} f(t)\} \). Each differentiation increases the power of \( t \) by one in the Laplace domain.
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

Solving Laplace Transform examples with differentiation property demonstrates its utility. Let's consider

• 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.

These examples underline how the differentiation property and integration by parts aid in not only calculating transforms but offering insights into the constructs of functions by leveraging the inherent relationships between time-domain and frequency-domain manipulations.