Question

Using integration by parts, find the Laplace transform of the given function; \(n\) is a positive integer and \(a\) is a real constant. $$ t^{2} \sinh a t $$

Short answer

Answer: The Laplace transform of the given function \(t^2 \sinh at\) is \(-\frac{2a}{(s^2 - a^2)(s+a)}\).

Step-by-step solution

Rewrite the given function with the Laplace definition

We need to rewrite the function \(t^2 \sinh at\) using the Laplace transform definition: $$ L[t^2 \sinh at] = \int_0^\infty e^{-st} \cdot t^2 \sinh at \, dt $$

Apply integration by parts

Let's assign some values for integration by parts: - Let \(u = t^2\), then \(du = 2tdt\) - Let \(dv = e^{-st} \sinh at dt\), then integrating each part with respect to \(t\) yields: - \(v = \int e^{-st} \sinh at dt\) Now we will use integration by parts: $$ \int_0^\infty e^{-st} \cdot t^2 \sinh at \, dt = \left[t^2 \cdot \int e^{-st} \sinh at dt \right]_0^\infty - \int_0^\infty 2t \cdot \left(\int e^{-st} \sinh at dt\right) dt $$

Find v using a known Laplace transform

We know the Laplace transform of \(\sinh at\) is \(\frac{a}{s^2 - a^2}\). So, let's find the Laplace transform of \(e^{-st} \sinh at dt\): $$ v = \int e^{-st} \sinh at dt = \frac{a}{s^2 - a^2} $$

Plug in v and simplify the equation

Plugging v back into the equation and simplifying: $$ \int_0^\infty e^{-st} \cdot t^2 \sinh at \, dt = \left[t^2 \cdot \frac{a}{s^2 - a^2} \right]_0^\infty - \int_0^\infty 2t \cdot \left(\frac{a}{s^2 - a^2}\right) dt =$$ $$ -\int_0^\infty \frac{2atd}{s^2 - a^2} dt $$

Integrate with respect to t

Now we need to integrate the remaining term with respect to \(t\): $$ -\int_0^\infty \frac{2atd}{s^2 - a^2} dt = -\frac{2a}{s^2 - a^2} \int_0^\infty td dt = -\frac{2a}{(s^2 - a^2)(s+a)} $$ We can now substitute this expression back into our original equation:

Write the final result

Our final answer for the Laplace transform of the given function is: $$ L[t^2 \sinh at] = -\frac{2a}{(s^2 - a^2)(s+a)} $$

Key concepts

Integration by Parts

In calculus, integration by parts is a technique used to integrate the product of two functions. According to the formula, \( \int u \frac{dv}{dt} dt = uv - \int v \frac{du}{dt} dt \), it allows us to break down complex integrals into simpler forms. This method is particularly useful when faced with a product of a polynomial and an exponential or trigonometric function, much like the exercise involving \( t^2 \sinh(at) \). By choosing \( u \) and \( dv \) strategically, we can reduce the original integral into a simpler form that can often be solved more directly.

For the exercise with \( t^2 \sinh(at) \), the choice of \( u \) and \( dv \) is key to finding a workable form. The process involves differentiating and integrating these parts to effectively use the integration by parts formula. Understanding and applying this concept can simplify the process of finding the Laplace transform, a common task in differential equations.

Laplace Transform of Sinh

The hyperbolic sine function, \(\sinh \), appears frequently in the study of differential equations and is related to the exponential functions. The Laplace transform of \(\sinh(at)\) is a standard result in calculus, given by \(\frac{a}{s^2 - a^2}\) as long as the real part of \(s\) is greater than the absolute value of \(a\). When solving for the Laplace transform of a function involving \(\sinh(at)\), such as in our exercise, this result simplifies the integration process substantially. By recognizing and utilizing the known Laplace transform of \(\sinh(at)\), we avoid the need to integrate it directly and can focus on integrating the remaining parts of the function.

Calculus

Calculus is the mathematical study of change, and it includes two major branches: differential and integral calculus. Integral calculus, which focuses on the accumulation of quantities and the areas under and between curves, directly applies to the process of finding Laplace transforms. In the context of our exercise, the techniques of calculus are essential to manipulate the integral that defines the Laplace transform, often employing the method of integration by parts. The goal is to express the given function, which may be complex and multifaceted, as a definite integral that can be evaluated to find the Laplace transform—translating a function from the time domain to the complex frequency domain.

Differential Equations

In mathematics, particularly in calculus, a differential equation is an equation that relates some function with its derivatives. These equations describe various phenomena in engineering, physics, economics, and biology. The Laplace transform is a powerful tool used to solve linear differential equations by converting the equation from a differential form in the time domain into an algebraic equation in the frequency domain. This transformation simplifies the process of finding solutions, especially when initial conditions are known. By applying the Laplace transform to both sides of a differential equation, we can often reduce the problem to an algebraic one, which is typically easier to solve. The original problem in our exercise is an example of finding a Laplace transform which, in a wider context, can be instrumental in solving differential equations.