Problem 21 - Differential Equations and Linear Algebra

Question

Determine the inverse Laplace transform of the given function. $$F(s)=\frac{2 s+3}{\left(s^{2}+4\right)\left(s^{2}+1\right)}.$$

Step-by-Step Solution

Answer

The inverse Laplace transform of the given function is: \(f(t) = \cos(2t) - 2\sin(2t) - \cos(t) + 2\sin(t)\)

Step 1: Partial Fraction Decomposition

We first need to decompose the given function into simpler fractions. The given function, \(F(s)\), can be rewritten as: \(F(s) = \frac{As + B}{s^2 + 4} + \frac{Cs + D}{s^2 + 1}\) Now, we need to find constants A, B, C, and D by equating the denominator and the numerator: \(\frac{2s + 3}{(s^2 + 4)(s^2 + 1)} = \frac{As + B}{s^2 + 4} + \frac{Cs + D}{s^2 + 1}\)

Step 2: Expanding the Equations and Equating Coefficients

Multiply both sides by \((s^2 + 4)(s^2 + 1)\) to clear the denominators. We get: \(2s + 3 = (As + B)(s^2 + 1) + (Cs + D)(s^2 + 4)\) Expanding this equation, we have: \(2s + 3 = As^3 + Bs^2 + As + B + Cs^3 + 4Cs + Ds^2 + 4D\) \(2s + 3 = (A+C)s^3 + (B+D)s^2 + (A+4C)s + (B+ 4D)\) Now, equate the coefficients of the corresponding powers of s: \((A+C) = 0\) (coefficient of \(s^3\)) \((B+D) = 0\) (coefficient of \(s^2\)) \((A + 4C) = 2\) (coefficient of s) \((B + 4D) = 3\) (constant term)

Step 3: Solving the Equations for A, B, C, and D

We now have a system of linear equations: \(A + C = 0\) \(B + D = 0\) \(A + 4C = 2\) \(B + 4D = 3\) Solving these equations (using substitution or matrix methods), we get: \(A = 1\), \(B = -2\), \(C = -1\), and \(D = 2\) Now, we can rewrite our original function as: \(F(s) = \frac{1s - 2}{s^2 + 4} + \frac{-1s + 2}{s^2 + 1}\)

Step 4: Inverse Laplace Transform of the Terms

Now, we can determine the inverse Laplace transform of each term: \(f(t) = \mathcal{L}^{-1}\{\frac{s-2}{s^2+4}\} + \mathcal{L}^{-1}\{\frac{-s+2}{s^2+1}\}\) Using the inverse Laplace Transform properties and tables, we find: \(f(t) = e^0 \cos(2t) - 2e^0\sin(2t) - e^0\cos(t) + 2e^0\sin(t)\)

Step 5: Simplify the Expression

Since all the terms have an exponent of 0, we can simplify the expression as: \(f(t) = \cos(2t) - 2\sin(2t) - \cos(t) + 2\sin(t)\) So, the inverse Laplace transform of the given function is: \(f(t) = \cos(2t) - 2\sin(2t) - \cos(t) + 2\sin(t)\)

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler components. This technique is particularly useful in the context of Laplace Transforms, as it allows us to work with simpler fractions when determining the inverse transform. In the given problem, the function \(F(s)\) is decomposed into fractions of the form \(\frac{As + B}{s^2 + 4} + \frac{Cs + D}{s^2 + 1}\). The purpose of this decomposition is to isolate terms that match standard Laplace Transform tables, making it easier to find the inverse transform later. To find the coefficients \(A\), \(B\), \(C\), and \(D\), we multiply through by the common denominator to equate the numerators. This entails expanding the expression and aligning like terms, ultimately leading to a system of linear equations to solve for the unknown constants. This step is crucial for rewriting the original function into a form suitable for further analysis.

Laplace Transform

The Laplace Transform is a mathematical operation that converts a function of time \(t\) into a function of a complex number \(s\). It is a powerful tool used in engineering and physics for analyzing linear time-invariant systems. The primary advantage of using the Laplace Transform is its ability to convert differential equations into algebraic equations, which are often easier to solve. In the given exercise, we start with a Laplace-transformed function \(F(s)\) and work towards finding its inverse. The purpose of the inverse Laplace Transform is to convert this function back into its time-domain counterpart, represented as \(f(t)\). By using the properties and tables associated with Laplace Transforms, we can map known \(s\)-domain functions back into the \(t\)-domain, allowing us to analyze the behavior of the original system.

System of Linear Equations

Solving a system of linear equations is a fundamental step in partial fraction decomposition and other mathematical methods involving algebraic manipulations. In this particular exercise, the coefficients \(A\), \(B\), \(C\), and \(D\) are determined by setting up a system of equations based on the decomposition of \(F(s)\). Here, we have four equations derived from equating the coefficients of different powers of \(s\):- \((A+C) = 0\)- \((B+D) = 0\)- \((A + 4C) = 2\)- \((B + 4D) = 3\)These equations can be solved using substitution or matrix methods, which provide a straightforward method for finding the unknowns. By solving this system, we determine the specific values of the coefficients needed to express \(F(s)\) in a decomposed form, which is critical for applying the inverse Laplace Transform.

Inverse Transform Properties

The inverse Laplace Transform is the method used to find a function in the time domain from its Laplace-transformed counterpart. In other words, we use this transform to convert data from the \(s\)-domain back to the \(t\)-domain. This provides insights into how a physical system behaves over time.In solving our present exercise, after applying partial fraction decomposition, we use the inverse Laplace Transform on each simplified fraction. Using standard transform tables and known properties, we compute:

• \(\mathcal{L}^{-1}\{\frac{s-2}{s^2+4}\}\) which results in terms involving \(\cos(2t)\) and \(\sin(2t)\).
• \(\mathcal{L}^{-1}\{\frac{-s+2}{s^2+1}\}\) which results in terms involving \(\cos(t)\) and \(\sin(t)\).

These inverse transformations are guided by key properties and the use of tables like the ones providing standard Laplace pairs. The results are combined to yield a single expression for \(f(t)\), which gives a clear picture of the system's response in the time domain.