Question

Calculate the Laplace transform of each of the following functions: (a) \(e^{-t} \cos 2 t\) (b) \(e^{-4 t} \cosh 2 t\) (c) \(\left(t^{2}+1\right)^{2}\) (d) \(3 \cosh t-4 \sinh 5 t\) (e) \(t^{n} \sin t\)

Short answer

(a) \(\frac{s+1}{s^2 + 2s + 5}\), (b) \(\frac{s+4}{s^2 + 8s + 12}\), (c) \(\frac{24}{s^5} + \frac{4}{s^3} + \frac{1}{s}\), (d) \(\frac{3s}{s^2 - 1} - \frac{20}{s^2 - 25}\), (e) \(\frac{n!}{(s^2+1)^{(n+1)}}\)

Step-by-step solution

Understand the Laplace Transform Definition

The Laplace Transform of a function \( f(t) \) is given by the integral: \( \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \,dt \). This transform is a tool for transforming differential equations into algebraic equations.

Find the Transform of (a) \( e^{-t} \cos 2t \)

Utilize the shifting property of the Laplace Transform, which states if \( f(t) \) has a transform \( F(s) \), then \( e^{at} f(t) \) has a transform \( F(s-a) \). The Laplace transform of \( \cos 2t \) is \( \frac{s}{s^2 + 4} \), so the transform of \( e^{-t} \cos 2t \) is \( \frac{s+1}{(s+1)^2 + 4} = \frac{s+1}{s^2 + 2s + 5} \).

Find the Transform of (b) \( e^{-4t} \cosh 2t \)

Recognize \( \mathcal{L}\{\cosh at\} = \frac{s}{s^2-a^2} \). Using the shifting property, \( \mathcal{L}\{ e^{-4t} \cosh 2t \} = \frac{s+4}{(s+4)^2 - 4} = \frac{s+4}{s^2 + 8s + 12} \).

Find the Transform of (c) \( (t^2 + 1)^2 \)

To find the Laplace transform of \( (t^2 + 1)^2 \), expand it first: \( t^4 + 2t^2 + 1 \). Use the linearity of the Laplace transform and standard transforms for \( t^n \): \( \frac{24}{s^5} + \frac{4}{s^3} + \frac{1}{s} \).

Find the Transform of (d) \( 3 \cosh t - 4 \sinh 5t \)

The Laplace transform for \( \cosh at \) and \( \sinh at \) are \( \frac{s}{s^2 - a^2} \) and \( \frac{a}{s^2 - a^2} \), respectively. Calculate separately and combine: \( \frac{3s}{s^2 - 1} - \frac{20}{s^2 - 25} \).

Find the Transform of (e) \( t^n \sin t \)

Use the Laplace Transform formula for \( t^n f(t) \): \( \frac{n!}{(s^2+1)^{(n+1)}} \). For \( t^n \sin t \), the transform is \( \frac{n!}{(s^2 + 1)^{(n+1)}} \). This form can be generalized for any integer \( n \).

Key concepts

Shifting Property

The shifting property of the Laplace Transform is an essential tool when dealing with exponential functions. It is particularly effective in simplifying the transformation of complex expressions. If you have a function \( f(t) \), and you know its Laplace Transform is \( F(s) \), the shifting property states:

• For \( e^{at} f(t) \), the Laplace Transform shifts by \( a \), becoming \( F(s-a) \).

This is helpful when working with terms like \( e^{-t} \) in an equation, as it allows the direct application of known transforms to modified arguments. In practice, once you know the Laplace Transform for a base function like \( \cos 2t \), you can readily apply the shifting property to tackle transformations involving a multiplication of the original function by an exponential term.
For instance, for \( e^{-t} \cos 2t \), the transform is adjusted from \( \frac{s}{s^2 + 4} \) to \( \frac{s+1}{s^2 + 2s + 5} \) due to the shift by \( -1 \). This makes handling exponential arguments far more manageable.

Hyperbolic Functions

Hyperbolic functions like \( \cosh \) and \( \sinh \) often appear in problems related to transformations. They are analogs of trigonometric functions and play a crucial role when solving differential equations and conducting transformations. The fundamental transforms are:

• For \( \cosh(at) \), it's \( \frac{s}{s^2-a^2} \).
• For \( \sinh(at) \), it's \( \frac{a}{s^2-a^2} \).

When encountering expressions like \( e^{-4t} \cosh 2t \), leveraging the shifting property alongside the basic hyperbolic transform allows us to simplify the problem, resulting in a straightforward expression: \( \frac{s+4}{s^2 + 8s + 12} \).
Hyperbolic functions often emerge in equations describing real-world phenomena like waves and oscillations, reinforcing the importance of mastering their transforms.

Linearity of Laplace Transform

The linearity property of the Laplace Transform is a potent feature that makes it a natural choice for analyzing linear systems and equations. This property indicates that the Laplace Transform of a sum of functions is the sum of the Laplace Transforms of each function, multiplied by any constants associated with them. In formula terms:

• For \( f(t) = a \, g(t) + b \, h(t) \), \( \mathcal{L}\{f(t)\} = a \, \mathcal{L}\{g(t)\} + b \, \mathcal{L}\{h(t)\} \).

Suppose we have a function such as \( (t^2 + 1)^2 \). We first expand it to \( t^4 + 2t^2 + 1 \). Each term is then transformed individually using known transforms for polynomials, combining them afterwards. This reduces a seemingly complex problem to smaller, more manageable parts, resulting in \( \frac{24}{s^5} + rac{4}{s^3} + rac{1}{s} \), which makes problem-solving in more extended scenarios both systematic and accessible.

Differential Equations

Laplace Transforms are extensively used to solve differential equations, as they convert differential equations into algebraic ones, simplifying the process. This is particularly useful when dealing with linear differential equations with constant coefficients.
When applying Laplace Transforms to differential equations, two core aspects are often leveraged:

• The ability to turn derivatives into multiplication. The nth order derivative \( f^{(n)}(t) \) transforms into \( s^n \mathcal{L}\{f(t)\} - s^{n-1}f(0) - \cdots\ - f^{(n-1)}(0) \).
• The capacity to add initial conditions directly, aiding in solving initial value problems seamlessly.

This transformation allows the original equation to be manipulated easily, making it possible to apply algebraic techniques for solutions, and then transform the result back to the time domain for interpretation. Thus, for students, understanding the link between differential equations and the Laplace Transform is crucial for solving real-world dynamic system problems effectively.