Question

Find the Laplace transform of each of the following expressions: (a) \((t+1)^{2}\) (b) \(\left(\mathrm{e}^{t}+t\right)^{2}\) (c) \(\frac{\sin 4 t}{2 \mathrm{e}^{3 t}}\) (d) \(2 \sin t \cos t\) (e) \(\frac{2 t^{2}}{3 \mathrm{e}^{2 t}}\)

Short answer

In summary, the Laplace transforms of the given expressions are: (a) \((t+1)^2 \Rightarrow \frac{2}{s^3}+\frac{1}{s^2}+\frac{1}{s}\) (b) \(\left(\mathrm{e}^{t}+t\right)^{2} \Rightarrow \frac{1}{s-2}+\frac{2}{(s-1)^2}+\frac{2}{s^3}\) (c) \(\frac{\sin 4 t}{2 e^{3 t}} \Rightarrow \frac{2}{(s+3)^2+16}\) (d) \(2\sin t\cos t \Rightarrow \frac{2}{s^2+4}\) (e) \(\frac{2 t^{2}}{3 e^{2 t}} \Rightarrow \frac{4}{3(s+2)^3}\)

Step-by-step solution

Laplace Transform Definition

The Laplace transform of a function f(t) is defined as: \(\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t)e^{-st} dt\) We will apply this definition to each expression. (a) \((t+1)^2\)

Apply Laplace Transform

Applying the Laplace transform to \((t+1)^2\): \(\mathcal{L}\{(t+1)^2\} = \int_{0}^{\infty} (t+1)^2e^{-st} dt\)

Expand and split the integral

Expanding \((t+1)^2\) and splitting the integral: \(\mathcal{L}\{(t+1)^2\} = \int_{0}^{\infty} (t^2 + 2t + 1)e^{-st} dt\) \(\mathcal{L}\{(t+1)^2\} = \int_{0}^{\infty} t^2e^{-st} dt+\int_{0}^{\infty} 2te^{-st} dt+\int_{0}^{\infty} e^{-st} dt\)

Solve the integrals

We will use integration by parts to solve the integrals: \(Laplace\ transform\ of\ t^{n\) }\(=\frac{n!}{s^{n+1}}\) So, \(\mathcal{L}\{(t+1)^2\} = \frac{2!}{s^3} +\frac{1!}{s^2}+\frac{0!}{s} =\frac{2}{s^3}+\frac{1}{s^2}+\frac{1}{s}\) (b) \(\left(\mathrm{e}^{t}+t\right)^{2}\)

Apply Laplace Transform

Applying the Laplace transform to \((e^t + t)^2\): \(\mathcal{L}\{(e^t+t)^2\} = \int_{0}^{\infty} (e^t+t)^2e^{-st} dt\)

Expand and split the integral

Expanding \((e^t+t)^2\) and splitting the integral: \(\mathcal{L}\{(e^t+t)^2\} = \int_{0}^{\infty} (e^{2t}+2te^t+t^2)e^{-st} dt\) \(\mathcal{L}\{(e^t+t)^2\} = \int_{0}^{\infty} e^{2t}e^{-st} dt+\int_{0}^{\infty} 2te^te^{-st} dt+\int_{0}^{\infty} t^2e^{-st} dt\)

Solve the integrals

Using properties of integrals and integration by parts: \(\mathcal{L}\{(e^t+t)^2\} = \frac{1}{(s-2)^1}+\frac{2}{(s-1)^2}+\frac{2}{s^3} = \frac{1}{s-2}+\frac{2}{(s-1)^2}+\frac{2}{s^3}\) (c) \(\frac{\sin 4 t}{2 e^{3 t}}\)

Apply Laplace Transform

Applying the Laplace transform to \(\frac{\sin(4t)}{2e^{3t}}\): \(\mathcal{L}\{\frac{\sin 4t}{2e^{3t}}\} = \frac{1}{2}\int_{0}^{\infty} \sin(4t)e^{-(s+3)t} dt\)

Solve the integral

Using properties of integrals: \(\mathcal{L}\{\frac{\sin 4t}{2e^{3t}}\} = \frac{1}{2}\frac{4}{(s+3)^2+4^2} =\frac{2}{(s+3)^2+16}\) (d) \(2\sin t\cos t\)

Apply Trigonometric Identity

First, simplify the expression using the identity \(2\sin t\cos t = \sin 2t\): \(2\sin t\cos t = \sin 2t\)

Apply Laplace Transform

Applying the Laplace transform to \(\sin 2t\): \(\mathcal{L}\{\sin 2t\} = \int_{0}^{\infty} \sin(2t)e^{-st} dt\)

Solve the integral

Using properties of integrals: \(\mathcal{L}\{\sin 2t\} = \frac{2}{s^2+2^2} =\frac{2}{s^2+4}\) (e) \(\frac{2 t^{2}}{3 e^{2 t}}\)

Apply Laplace Transform

Applying the Laplace transform \(\frac{2t^2}{3e^{2t}}\): \( \mathcal{L}\{\frac{2 t^{2}}{3 e^{2 t}}\} = \frac{2}{3}\int_{0}^{\infty} t^{2}e^{-(s+2)t} dt\)

Solve the integral

Using integration by parts: \(\mathcal{L}\{\frac{2 t^{2}}{3 e^{2 t}}\}= \frac{2}{3} \frac{2!}{(s+2)^3} = \frac{4}{3(s+2)^3}\)

Key concepts

Integration by Parts

Integration by parts is an essential mathematical technique used to integrate products of functions. It's based on the product rule for differentiation and can often simplify complex integrals. The rule is given by
\[ \int u dv = uv - \int v du \]
where one part of the product (\( u \)) is differentiated (\( du \)) and the other (\( dv \)) is integrated to find \( v \). Choosing \( u \) and \( dv \) wisely is crucial, as the goal is to make the resulting integral \( \int v du \) simpler to solve.
For instance, in the exercise, when integrating \( t^2 e^{-st} \), by letting \( u = t^2 \), and \( dv = e^{-st} dt \), we can find \( du \) and \( v \) and apply the technique to obtain the transformed function in terms of \( s \). This approach greatly aids in finding the Laplace transform of polynomial functions.

Laplace Transform of Functions

The Laplace transform is a powerful tool used to convert functions of time (\( t \)) into functions of a complex variable (\( s \)). This transformation, shown as \( \mathcal{L}\{f(t)\} = F(s) \), transforms the time domain into the frequency or complex domain. It is particularly useful in solving differential equations and analyzing systems in engineering and physics.
Applying the Laplace transform, as you've seen in the solved exercise examples, involves integrating the function \( f(t) \) multiplied by an exponential decay factor \( e^{-st} \) from zero to infinity. The result is a function of \( s \) which can be manipulated more easily than the original time function. This method not only simplifies calculations but also provides insight into the behavior of the original system represented by \( f(t) \).

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They are incredibly useful in simplifying expressions and solving equations. The exercise part (d) showcases the use of a fundamental trigonometric identity: \( 2\sin(t)\cos(t) = \sin(2t) \).
This specific identity is known as the double angle formula and it's invaluable in transforming products of sine and cosine into a single trigonometric function. With this identity, integrating the transformed function becomes straightforward, illustrating how recognizing and applying trigonometric identities can vastly simplify the process of finding the Laplace transform of trigonometric expressions.

Exponential Function

The exponential function is pivotal in both mathematics and the sciences, characterized by the equation \( e^x \), where \( e \) is the base of the natural logarithm. Its relevance is highlighted in its relationship with growth and decay processes.
In our context of Laplace transforms, it's the exponential decay factor \( e^{-st} \) that plays a key role. This factor ensures that the integral converges for functions that otherwise would diverge. Moreover, the behavior of the exponential function allows us to examine stability and transient behaviors in physical systems.
For example, in the Laplace transform of \( (e^t + t)^2 \), the presence of the term \( e^{2t} e^{-st} \) highlights this function's utility, showing how exponential growth and decay are intertwined in analysis and solutions in the frequency domain.