Problem 44 - Precalculus

Question

Refer to the following: Laplace transforms are used to solve differential equations. The Laplace transform of \(f(t)\) is denoted by \(L\\{f(t)\\} ;\) thus, \(L\left\\{e^{3 t}\right\\}\) is the Laplace transform of \(f(t)=e^{3 t} .\) It is known that \(L\left\\{e^{k t}\right\\}=\frac{1}{s-k}\) and \(L\left\\{e^{-k t}\right\\}=\frac{1}{s+k} .\) Then the inverse Laplace transform of \(g(s)=\frac{1}{s-k}\) is \(L^{-1}\left\\{\frac{1}{s-k}\right\\}=e^{k t} .\) Inverse Laplace transforms are linear: $$L^{-1}\\{f(t)+g(t)\\}=L^{-1}\\{f(t)\\}+L^{-1}\\{g(t)\\}$$ Use partial fractions to find the inverse Laplace transform of \(\frac{2 s^{2}+3 s-2}{s(s+1)(s-2)}\).

Step-by-Step Solution

Answer

The inverse Laplace transform is \( 1 + 2e^{-t} - e^{2t} \).

Step 1: Perform Partial Fraction Decomposition

The expression \( \frac{2s^2 + 3s - 2}{s(s+1)(s-2)} \) must be decomposed into simpler fractions. We assume a form: \( \frac{2s^2 + 3s - 2}{s(s+1)(s-2)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s-2} \).

Step 2: Solve for Constants A, B, and C

Multiply both sides by \( s(s+1)(s-2) \) to eliminate the denominators, arriving at the equation: \( 2s^2 + 3s - 2 = A(s+1)(s-2) + Bs(s-2) + Cs(s+1) \). Next, expand and collect like terms to solve for \( A, B, \) and \( C \) by substituting convenient values for \( s \) or by equating coefficients. Substituting \( s = 0, -1, \) and \( 2 \) yields: \( A = 1, B = 2, C = -1 \).

Step 3: Write the Decomposed Expression

Using the values found for \( A, B, \) and \( C \), express the partial fraction decomposition: \( \frac{2s^2 + 3s - 2}{s(s+1)(s-2)} = \frac{1}{s} + \frac{2}{s+1} - \frac{1}{s-2} \).

Step 4: Find the Inverse Laplace Transform of Each Term

Apply the inverse Laplace transform to each separate term using known formulas:1. \( L^{-1}\{\frac{1}{s}\} = 1 \), since \( \mathcal{L}^{-1}\{\frac{1}{s}\} = 1 \).2. \( L^{-1}\{\frac{2}{s+1}\} = 2e^{-t} \), since \( \mathcal{L}^{-1}\{\frac{1}{s+a}\} = e^{-at} \).3. \( L^{-1}\{\frac{-1}{s-2}\} = -e^{2t} \), since \( \mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{at} \).

Step 5: Combine the Results

By using the linearity of the inverse Laplace transform, combine the results from Step 4:- \( L^{-1}\left\{\frac{2s^2 + 3s - 2}{s(s+1)(s-2)}\right\} = 1 + 2e^{-t} - e^{2t} \).

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used to break down complex rational expressions into a sum of simpler fractions. This process is particularly useful in calculus, especially when dealing with Laplace transforms, as it simplifies the expressions, making them easier to manipulate and integrate.
To perform partial fraction decomposition, you start by expressing your complex fraction in terms of its simplest components. For instance, in our example, the fraction \( \frac{2s^2 + 3s - 2}{s(s+1)(s-2)} \) is rewritten as \( \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s-2} \).
By multiplying through by the common denominator, you eliminate the fractions and equate the resulting polynomial. This helps in determining the values of \( A, B, \) and \( C \) either by choosing strategic values for \( s \) or by matching coefficients.

• Set \(s\) to convenient values to solve for \(A, B, \) and \(C\).
• For example, by substituting different special values like \(s = 0, -1,\ and \ 2\), you can isolate and find the coefficients.

Finding these constants allows you to express the original fraction in a decomposed form, which is a crucial step when applying inverse Laplace transforms.

Inverse Laplace Transform

The inverse Laplace transform is a technique used to find the original function from its Laplace transform. It's a critical tool in solving differential equations, especially when initial conditions are provided.
With the inverse Laplace transform, you essentially reverse the process of the Laplace transform, translating a function of frequency back into a function of time. When dealing with partial fractions, the task becomes applying known inverse transforms to each decomposed part.

• For each item like \( \frac{1}{s}, \frac{2}{s+1}, \) and \( \frac{-1}{s-2} \), apply the inverse Laplace:
• \( L^{-1}\{\frac{1}{s}\} = 1\)
• \( L^{-1}\{\frac{2}{s+1}\} = 2e^{-t}\)
• \( L^{-1}\{\frac{-1}{s-2}\} = -e^{2t}\)

By computing these inverse transformations, you convert the solution back into the time domain, yielding a more understandable differential solution.

Differential Equations

Differential equations are mathematical equations that involve unknown functions and their derivatives. They are a powerful tool for modeling various phenomena in engineering, physics, and other sciences.
Using the Laplace transform, complex differential equations can be transformed into simpler algebraic problems.
The Laplace transform operates under the integral transform technique, which helps convert differential equations into algebraic ones. This simplification makes finding the solution more straightforward, especially in equations with complex initial conditions.

• The differential equations tackled through the Laplace transform generally exhibit linearity, making them simpler to solve.
• The solution to the transformed equation can be found via inverse Laplace, reverting to the function of time.

This capability of translating between domains sets the Laplace transform apart as a valuable method in solving real-world differential equations.

Linear Operations

In mathematics, linearity is a property that describes the ability to break functions, equations, or transformations into simpler constituent parts. When dealing with Laplace transforms, linear operations allow these processes to be applied and simplified extensively.
The linearity of the Laplace transform and its inverse allows us to manipulate and individually solve parts of more extensive expressions. For instance, when you decompose a complex expression into partial fractions, each fraction can be separately inversed.

• Linear operations are reflected in the additive property: \(L^{-1}\{f(t) + g(t)\} = L^{-1}\{f(t)\} + L^{-1}\{g(t)\}\).
• This property makes solving for inverse Laplace transforms easier—each term is handled independently and summed for the full solution.

In contexts like differential equations or signal processing, this linear approach facilitates efficient, modular computation, emphasizing the importance of linear operations.