Question

Find the inverse Laplace transform of:\(\frac{p}{(p+a)^{3}}\)

Short answer

te^{-at} - \frac{a}{2} t^{2} e^{-at}

Step-by-step solution

Identify the given Laplace Transform

The given Laplace transform is \(\frac{p}{(p+a)^{3}}\)

Express in known Laplace pairs

We need to express \(\frac{p}{(p+a)^{3}}\) in terms of known Laplace transform pairs. Recognize that this is related to the known Laplace transform pair for \(\frac{1}{(p+a)^{n}}\), which corresponds to \(\frac{t^{n-1}}{(n-1)!} e^{-at}\).

Decompose the given transform

Rewrite \(\frac{p}{(p+a)^{3}}\) as \(\frac{(p+a) - a}{(p+a)^{3}} = \frac{p+a}{(p+a)^{3}} - \frac{a}{(p+a)^{3}} = \frac{1}{(p+a)^{2}} - \frac{a}{(p+a)^{3}}\).

Use the Laplace inverse pairs

Apply the known inverse Laplace transform pairs: - The Laplace inverse of \(\frac{1}{(p+a)^{2}}\) is \({te^{-at}}\). - The Laplace inverse of \(\frac{a}{(p+a)^{3}}\) is \(\frac{a}{2!}t^{2}e^{-at}\).

Write the complete inverse

Combine both results to obtain the full inverse Laplace transform: \[\begin{equation} \mathcal{L}^{-1}\bigg(\frac{p}{(p+a)^{3}}\bigg) = te^{-at} - \frac{a}{2} t^{2} e^{-at} \end{equation}\].

Key concepts

Laplace transform pairs

To understand how we solve inverse Laplace transform problems, you need to be familiar with Laplace transform pairs. These pairs are fundamental because they connect functions in the s-domain (Laplace domain) with their counterparts in the time domain (t-domain).
For instance, the Laplace transform of \(\frac{1}{(p+a)^n}\) is known to be \(\frac{t^{n-1}}{(n-1)!} e^{-at}\). Knowing these pairs allows us to quickly identify how to revert to the original time function.
In simpler words, once we recognize a pattern in the transform function, we compare it with these known pairs to find its inverse. This comparison is the first and crucial step in finding the inverse Laplace transform.

partial fraction decomposition

Partial fraction decomposition is a technique used to break down complex fractions into simpler ones. These simpler fractions can then be easily managed or transformed.
Let’s look at the exercise you provided: We start with the fraction \(\frac{p}{(p+a)^{3}}\). By expressing this in a simpler form, we rewrite it using partial fractions.
Specifically, we break it down as follows:
\(\frac{p}{(p+a)^3} = \frac{(p+a) - a}{(p+a)^3} \). This simplifies to: \(\frac{p+a}{(p+a)^3} - \frac{a}{(p+a)^3} = \frac{1}{(p+a)^2} - \frac{a}{(p+a)^3} \).
This decomposition is key. It allows us to apply the inverse Laplace transform directly, using known pairs.

inverse Laplace operations

Inverse Laplace operations convert a given Laplace transform back into its original time-domain function. Once the fraction is simplified using known Laplace pairs, we apply the inverse operation.
From the simplified fractions in our exercise:
- \(\frac{1}{(p+a)^{2}}\) corresponds to \(te^{-at}\).
- \(\frac{a}{(p+a)^{3}}\) relates to \(\frac{a}{2!}t^{2}e^{-at}\).
By identifying these terms, we revert them back to their time functions. This step involves both recognition of the Laplace pairs and confidence in applying those pairs. It's helpful to think of this as reversing the original process, systematically returning from the s-domain to the t-domain.

differential equations

Inverse Laplace transforms are a powerful tool in solving differential equations. They can simplify complex linear differential equations by transforming them into the s-domain, solving algebraically, and then transforming back.
Let’s apply this concept: Suppose you have a differential equation in the time-domain that's complex to solve directly. By taking the Laplace transform, this equation becomes simpler. Once solved in the s-domain, we use the inverse Laplace transform to revert back.
The inverse Laplace transform is integral in numerous fields, including engineering and physics, to solve problems with exponential growth or decay, oscillatory systems, and more.