Question

Find the inverse Laplace transform of the given function. $$ \frac{3 s}{s^{2}-s-6} $$

Short answer

Answer: The inverse Laplace transform of the function \(\frac{3s}{s^2 - s - 6}\) is \(\frac{6}{5}e^{-2t}+\frac{9}{5}e^{3t}\).

Step-by-step solution

Perform Partial Fraction Decomposition

First, we need to factor the denominator of the given function: $$ s^{2} - s - 6 = (s+2)(s-3) $$ Now, we can perform partial fraction decomposition on the given function: $$ \frac{3s}{(s+2)(s-3)} = \frac{A}{s+2} + \frac{B}{s-3} $$ To find the constants A and B, we can clear the fractions by multiplying both sides of the equation by \((s+2)(s-3)\): $$ 3s = A(s-3) + B(s+2) $$

Solve for the Constants A and B

Now we need to find the values of A and B. We can do this by solving the equation \(3s = A(s-3) + B(s+2)\) for different values of s. To find A, we can let s = 3: $$ 3(3) = A(3-3) + B(3+2) \\ 9 = 5B \\ B = \frac{9}{5} $$ To find B, we can let s = -2: $$ 3(-2) = A(-2-3) + B(-2+3) \\ -6 = -5A \\ A = \frac{6}{5} $$

Rewrite the Function with A and B

Now that we have the values of A and B, we can rewrite the function as: $$ \frac{3s}{(s+2)(s-3)} = \frac{\frac{6}{5}}{s+2} + \frac{\frac{9}{5}}{s-3} $$

Use Laplace Transform Table to Find Inverse Laplace Transform

Now we can use the Laplace transform table to find the inverse Laplace transform of each term: $$ \mathcal{L}^{-1} \left\{\frac{\frac{6}{5}}{s+2}\right\} = \frac{6}{5} \mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\} = \frac{6}{5} e^{-2t} \\ \mathcal{L}^{-1} \left\{\frac{\frac{9}{5}}{s-3}\right\} = \frac{9}{5} \mathcal{L}^{-1}\left\{\frac{1}{s-3}\right\} = \frac{9}{5} e^{3t} $$ Finally, we can sum the two transformed terms to find the inverse Laplace transform of the original function: $$ \mathcal{L}^{-1}\left\{\frac{3s}{(s+2)(s-3)}\right\} = \frac{6}{5} e^{-2t} + \frac{9}{5} e^{3t} $$ So the inverse Laplace transform of the given function is: $$ \frac{6}{5} e^{-2t} + \frac{9}{5} e^{3t} $$

Key concepts

Partial Fraction Decomposition

When working with complex rational functions, particularly when trying to find their inverse Laplace transforms, partial fraction decomposition is an invaluable tool. This technique involves breaking down a complex fraction into simpler components called 'partial fractions,' which can be more easily transformed or integrated.

The process starts with factorizing the denominator of the original function. Once the denominator is factorized, the fraction is expressed as a sum of terms where each term has one of the factors in its denominator and an unknown constant in its numerator. These constants are what we seek to determine in order to complete the decomposition.

Solving this set-up usually requires setting up an equation by multiplying both sides by the denominator, thus clearing the fractions. You then solve for the unknown constants by substituting strategic values for the variable that simplify the equation or by equating coefficients of like terms. In our example, we substituted the roots of the denominator to quickly find the constants.

The correctly decomposed form enables easier computation when dealing with integral transforms like the Laplace transform, and is crucial for handling more complex differential equations.

Laplace Transform Table

A Laplace transform table is an indispensable reference in the realm of differential equations and control systems, listing many transforms of common functions. The table allows us to directly obtain the Laplace transform or its inverse without going through the laborious process of integration that defines the Laplace transform.

For each entry in the table, there is a function in the time domain and its counterpart in the Laplace or frequency domain. For instance, the exponential function \(e^{at}\) has a Laplace transform of \(\frac{1}{s-a}\) if we're looking at the unilateral transform.

Once we've decomposed a function using partial fractions, these tables allow us to write down the inverse transforms from memory or by quick reference, rather than performing complex integrations. With the constants A and B identified, and the corresponding standard forms located in the table, we can precisely map our function back to the time domain, simplifying our solution considerably.

Solving Linear Equations

Linear equations form the backbone of a wide range of mathematical problems, including the process of finding constants in partial fraction decomposition. These equations, which can represent lines in two-dimensional space, are vital in describing relationships where each term is either a constant or the product of a constant and a single variable.

To solve linear equations, one might isolate the variable or equate coefficients from both sides of the equation. This commonly involves addition, subtraction, multiplication or division to manipulate the equation into a more manageable form. In the context of our example with partial fraction decomposition, we strategically choosen values for the variable that simplify the equation, allowing for the straightforward isolation of our unknown constants.

The knowledge of solving linear equations is fundamental. It allows students to progress into more complex areas such as calculus, linear algebra, and beyond, and is practiced frequently in engineering fields, where systems are often modeled or controlled using linear relationships.