Problem 29 - Differential Equations and Linear Algebra

Question

Determine \(L^{-1}[F]\). $$F(s)=\frac{4}{(s+2)^{3}}$$.

Step-by-Step Solution

Answer

The inverse Laplace transform of the given function \(F(s) = \frac{4}{(s+2)^3}\) is \(L^{-1}[F(s)] = t^2 e^{-2t}\).

Step 1: Identify Known Laplace Transform Pairs

We begin by identifying any known Laplace transform pairs that might help us solve the problem. From the Laplace transform table, we find the following pair relates to the given function: \[ L^{-1}\bigg[\frac{n!}{(s+a)^{n+1}}\bigg] = t^n e^{-at} \] This pair will be useful as the given function has a similar form.

Step 2: Rewrite the Given Function

Now we rewrite the given function, \(F(s)\), using the known pair's format so that we can apply the inverse Laplace transform directly. We have: \[ F(s) = \frac{4}{(s+2)^3} = \frac{2!}{(s+2)^3}, \quad \text{where } n = 2, a = 2. \]

Step 3: Apply the Inverse Laplace Transform Formula

Using the known inverse Laplace transform pair, we find that the inverse Laplace transform of \(F(s)\) is: \[ L^{-1}[F(s)] = t^2 e^{-2t} \] Thus, the inverse Laplace transform of the given function, denoted as \(L^{-1}[F(s)]\), is \(t^2 e^{-2t}\).

Key Concepts

Inverse Laplace Transform

The Inverse Laplace Transform is a core concept in engineering and mathematics. It allows us to move from the Laplace domain, often represented in terms of variable \( s \), back into the time domain, usually in terms of variable \( t \). This is crucial for analyzing real-world systems, especially in fields like electrical engineering and control systems.

• The inverse process is often about finding a function of \( t \) from its Laplace-transformed version in \( s \).
• This involves recognizing patterns and using known transform pairs.
• It allows us to solve differential equations more easily as they become algebraic in the \( s \)-domain.

In the problem we discussed, the Laplace transform of a specific time-domain function, \( t^2 e^{-2t} \), shows up as \( \frac{4}{(s+2)^3} \) in the \( s \)-domain.
By applying the inverse Laplace transform, we effectively "undo" the transformation, thus finding \( t^2 e^{-2t} \), which gives us back the original time-based function.

Laplace Transform Pairs

Laplace Transform Pairs are essential tools that make the process of finding inverse transforms much simpler.

• They are pairs of functions where one is in time-domain \( t \) and the other in the Laplace domain \( s \).
• These pairs are usually found in a table of transforms and inverses, which acts as a reference guide.
• By matching the function in the Laplace domain with its pair, we can easily find the corresponding time-domain function.

In our example, the function \( F(s) = \frac{4}{(s+2)^3} \) corresponds to the pair \( L^{-1}\big[\frac{n!}{(s+a)^{n+1}}\big] = t^n e^{-at} \).
From here, we matched parameters to identify that \( n = 2 \), and \( a = 2 \), leading us directly to the solution \( t^2 e^{-2t} \).
This dramatically simplifies solving inverse transforms without direct computation.

Mathematical Problem Solving

The process of mathematical problem-solving using inverse Laplace transforms often involves several logical steps.

• Identify: Start by identifying known transform pairs from a reference.
• Rewrite: Transform the given \( F(s) \) into a form that pairs can easily be applied to.
• Apply: Utilize the identified pair to find its time-domain equivalent swiftly.

These steps ensure a smooth transition from the mathematical abstraction in the \( s \)-domain back to physical reality in the \( t \)-domain.
In the example given, rewriting \( \frac{4}{(s+2)^3} \) as \( \frac{2!}{(s+2)^{3}} \) allowed us to see that it corresponds nicely to a form in the transform pairs table.
Applying the pair inversely gives the solution in real terms, \( t^2 e^{-2t} \), showcasing the efficiency and power of using Laplace transform methods to tackle complex differential equations or signal processing problems.