Question

Find the inverse Laplace transform. $$F(s)=\frac{2 s-3}{s^{2}-3 s+2}$$

Short answer

In this exercise, we found the inverse Laplace transform of the function $$F(s)=\frac{2 s-3}{s^{2}-3 s+2}$$. We used partial fraction decomposition to first rewrite the function as a sum of simpler fractions: $$F(s) = -\frac{1}{s-1} + \frac{1}{s-2}$$. Then, we applied the inverse Laplace transform to each term, resulting in the time-domain function $$f(t) = - e^{t} + e^{2t}$$.

Step-by-step solution

Perform Partial Fraction Decomposition

We first rewrite the function as a sum of simpler fractions. We can decompose the given function into: $$\frac{2 s-3}{s^{2}-3 s+2} = \frac{A}{s-1} + \frac{B}{s-2}$$ To solve for A and B, we will multiply both sides by the denominator to clear the fractions: $$2s-3 = A(s-2) + B(s-1)$$ Now, let's solve for A and B by choosing appropriate values for s: 1. Let s = 1, we get: $$2(1)-3 = A(1-2) \Rightarrow A = -1$$ 2. Let s = 2, we get: $$2(2)-3 = B(2-1) \Rightarrow B = 1$$ Now we have the partial fractions: $$F(s) = \frac{-1}{s-1} + \frac{1}{s-2}$$

Apply the Inverse Laplace Transform

Now we can apply the inverse Laplace transform to each term. Remember that: $$\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{a t}$$ Apply the inverse Laplace transform to each term: $$f(t) = \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1} \{ -\frac{1}{s-1} + \frac{1}{s-2} \}$$ This gives us: $$f(t) = - e^{(1)t} + e^{(2)t}$$ So, the inverse Laplace transform of the given function $$F(s)=\frac{2 s-3}{s^{2}-3 s+2}$$ is: $$f(t) = - e^{t} + e^{2t}$$

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to work with. In the given exercise, we need to decompose the expression \( \frac{2s-3}{s^{2}-3s+2} \). This involves expressing it as a sum of fractions with linear denominators, such as \( \frac{A}{s-1} + \frac{B}{s-2} \).To find the coefficients \(A\) and \(B\), we clear the fractions by multiplying through by the original denominator. We use strategic values for \(s\) that simplify the equation. Choosing \(s=1\) and \(s=2\) quickly helps us find \(A=-1\) and \(B=1\). Now, the original complex fraction is rewritten as \(\frac{-1}{s-1} + \frac{1}{s-2}\). This makes it much easier to apply further analysis, such as the inverse Laplace transform.

Laplace Transform Properties

The Laplace transform is a powerful tool in mathematical analysis, especially useful in solving differential equations and systems dynamics. Its properties include linearity, frequency shifting, and others that make it incredibly versatile in engineering and physics problems.In this exercise, we focus on the inverse Laplace transform of the decomposed function \(F(s) = \frac{-1}{s-1} + \frac{1}{s-2}\). Utilizing the property \( \mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{a t} \), we can take each fraction separately to transform back into the time domain. This transform is crucial for interpreting solutions in the context where they were initially presented, such as physical processes or signal processing tasks. By backward transforming each component, we efficiently arrive at the solution: \( f(t) = -e^{t} + e^{2t} \).

Differential Equations

Differential equations describe how quantities change and are fundamental in mathematical modeling of real-world systems. They often appear in physics, engineering, biology, and finance. Solving differential equations can sometimes require transforming them into simpler forms, using methods such as the Laplace transform.In the exercise provided, finding the inverse Laplace transform is equivalent to solving a differential equation whose solution is expressed as a function of time, \(f(t)\). Once we decompose the original expression into simpler parts, we successfully leverage the inverse Laplace transform to directly find the temporal responses \(-e^{t} + e^{2t}\). This solution reflects how the initial conditions or inputs to the system behave over time, an insight critical for engineers and scientists.

Mathematical Analysis

Mathematical analysis provides a foundational framework for understanding concepts like limits, continuity, and transformations. It rigorously deals with the convergence of sequences and functions, ensuring solutions are mathematically sound.In finding inverse Laplace transforms, we engage in mathematical analysis by verifying that the breakdown and transformations we apply preserve the integrity of the solution. Every step, from partial fraction decomposition to final transformation, involves considering limits and continuity. This precision helps ensure our inferred time-domain solution, \( f(t) = -e^{t} + e^{2t} \), accurately reflects the original problem statement and adheres to all necessary mathematical conditions. Thus, mathematical analysis not only guides the solution process but ensures it is robust and reliable for theoretical or practical application.