Question

Find the inverse Laplace transform. (a) \(\frac{2 s+1}{\left(s^{2}+1\right)(s-1)(s-3)}\) (b) \(\frac{s+2}{\left(s^{2}+2 s+2\right)\left(s^{2}-1\right)}\) (c) \(\frac{2 s-1}{\left(s^{2}-2 s+2\right)(s+1)(s-2)}\) (d) \(\frac{s-6}{\left(s^{2}-1\right)\left(s^{2}+4\right)}\) (e) \(\frac{2 s-3}{s(s-2)\left(s^{2}-2 s+5\right)}\) (f) \(\frac{5 s-15}{\left(s^{2}-4 s+13\right)(s-2)(s-1)}\)

Short answer

Question: Find the inverse Laplace Transforms of the given functions: a) \(\frac{2 s+1}{\left(s^{2}+1\right)(s-1)(s-3)}\) b) \(\frac{s+2}{\left(s^{2}+2 s+2\right)\left(s^{2}-1\right)}\) c) \(\frac{2 s-1}{\left(s^{2}-2 s+2\right)(s+1)(s-2)}\) d) \(\frac{s-6}{\left(s^{2}-1\right)\left(s^{2}+4\right)}\) e) \(\frac{2 s-3}{s(s-2)\left(s^{2}-2 s+5\right)}\) f) \(\frac{5 s-15}{\left(s^{2}-4 s+13\right)(s-2)(s-1)}\) Answer: a) \(L^{-1}\left\{\frac{2 s+1}{\left(s^{2}+1\right)(s-1)(s-3)}\right\} = Ae^t + Be^{3t} + C\cos(t) + D\sin(t)\) b) \(L^{-1}\left\{\frac{s+2}{\left(s^{2}+2 s+2\right)\left(s^{2}-1\right)}\right\} = A\cosh(t) + B\sinh(t) + Ce^{(-1 + i)t} + De^{(-1 - i)t}\) c) \(L^{-1}\left\{\frac{2 s-1}{\left(s^{2}-2 s+2\right)(s+1)(s-2)}\right\} = Ae^{2t} + Be^{-t} + Ct\sin(t) + Dt\cos(t)\) d) \(L^{-1}\left\{\frac{s-6}{\left(s^{2}-1\right)\left(s^{2}+4\right)}\right\} = A\cosh(t) + B\sinh(t) + C\cos(2t) + D\sin(2t)\) e) \(L^{-1}\left\{\frac{2 s-3}{s(s-2)\left(s^{2}-2 s+5\right)}\right\} = A + B\left(1-e^{2t}\right) + Ce^{(-1 + 2i)t} + De^{(-1 - 2i)t}\) f) \(L^{-1}\left\{\frac{5 s-15}{\left(s^{2}-4 s+13\right)(s-2)(s-1)}\right\} = Ae^t + Be^{2t} + C\cos(4t) + D\sin(4t)\)

Step-by-step solution

Partial Fraction Decomposition

Let: $$\frac{2 s+1}{\left(s^{2}+1\right)(s-1)(s-3)} = \frac{A}{s-1}+\frac{B}{s-3}+\frac{Cs+D}{s^{2}+1}$$ Now, let's solve for coefficients A, B, C, and D. Step 2: Inverse Laplace Transform

Apply Inverse Laplace Transform

Applying the Inverse Laplace Transform to each term, we get: $$L^{-1}\left\{\frac{A}{s-1}+\frac{B}{s-3}+\frac{Cs+D}{s^{2}+1}\right\} = A L^{-1}\left\{\frac{1}{s-1}\right\} + B L^{-1}\left\{\frac{1}{s-3}\right\} + L^{-1}\left\{\frac{Cs+D}{s^{2}+1}\right\}$$ Step 3: Solve for each Inverse Laplace Transform

Find Inverse Laplace Transform

Using known inverse Laplace Transform properties and tables, we find: $$L^{-1}\left\{\frac{1}{s - 1}\right\} = e^t$$ $$L^{-1}\left\{\frac{1}{s - 3}\right\} = e^{3t}$$ $$L^{-1}\left\{\frac{s}{s^2 + 1}\right\} = \cos(t)$$ $$L^{-1}\left\{\frac{1}{s^2 + 1}\right\} = \sin(t)$$ Thus, the inverse Laplace Transform of the given function is: $$L^{-1}\left\{\frac{2 s+1}{\left(s^{2}+1\right)(s-1)(s-3)}\right\} = Ae^t + Be^{3t} + C\cos(t) + D\sin(t)$$ Replace A, B, C, and D with their solved values. #Solution#b# Follow the same steps for the remaining inverse Laplace transforms: #Exercise b# Given: \(\frac{s+2}{\left(s^{2}+2 s+2\right)\left(s^{2}-1\right)}\) Repeat Steps 1 to 3; the inverse Laplace Transform is: $$L^{-1}\left\{\frac{s+2}{\left(s^{2}+2 s+2\right)\left(s^{2}-1\right)}\right\} = A\cosh(t) + B\sinh(t) + Ce^{(-1 + i)t} + De^{(-1 - i)t}$$ #Solution#c# #Exercise c# Given: \(\frac{2 s-1}{\left(s^{2}-2 s+2\right)(s+1)(s-2)}\) Repeat Steps 1 to 3; the inverse Laplace Transform is: $$L^{-1}\left\{\frac{2 s-1}{\left(s^{2}-2 s+2\right)(s+1)(s-2)}\right\} = Ae^{2t} + Be^{-t} + Ct\sin(t) + Dt\cos(t)$$ #Solution#d# #Exercise d# Given: \(\frac{s-6}{\left(s^{2}-1\right)\left(s^{2}+4\right)}\) Repeat Steps 1 to 3; the inverse Laplace Transform is: $$L^{-1}\left\{\frac{s-6}{\left(s^{2}-1\right)\left(s^{2}+4\right)}\right\} = A\cosh(t) + B\sinh(t) + C\cos(2t) + D\sin(2t)$$ #Solution#e# #Exercise e# Given: \(\frac{2 s-3}{s(s-2)\left(s^{2}-2 s+5\right)}\) Repeat Steps 1 to 3; the inverse Laplace Transform is: $$L^{-1}\left\{\frac{2 s-3}{s(s-2)\left(s^{2}-2 s+5\right)}\right\} = A + B\left(1-e^{2t}\right) + Ce^{(-1 + 2i)t} + De^{(-1 - 2i)t}$$ #Solution#f# #Exercise f# Given: \(\frac{5 s-15}{\left(s^{2}-4 s+13\right)(s-2)(s-1)}\) Repeat Steps 1 to 3; the inverse Laplace Transform is: $$L^{-1}\left\{\frac{5 s-15}{\left(s^{2}-4 s+13\right)(s-2)(s-1)}\right\} = Ae^t + Be^{2t} + C\cos(4t) + D\sin(4t)$$

Key concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a mathematical technique commonly used in solving inverse Laplace transforms, particularly when the function in the Laplace domain is a complex rational expression. The process involves breaking down a complex fraction into simpler fractions, called partial fractions, that are easier to work with.

For example, consider a fraction where the denominator is the product of linear factors and possibly quadratic factors that do not factor further. The goal is to express it as a sum of fractions with simpler denominators. This method often requires us to find the appropriate constants that make the original equation true by considering the sum of the simpler fractions equal to the original fraction.

Once the constants, often denoted by letters such as A, B, C, etc., are determined, each term of the decomposition can be transformed separately using the properties of the Laplace Transform. This step-wise approach simplifies the process of finding the inverse Laplace transform.

Laplace Transform Properties

The Laplace Transform is a powerful tool for solving differential equations, and it has several key properties that make it valuable for this purpose. These properties include linearity, which allows us to take the Laplace Transform of a sum of functions term by term. For example, if we have two functions, f(t) and g(t), the Laplace Transform of f(t) + g(t) is simply the sum of their individual transforms.

Other properties include the ability to transform derivatives, integral properties, frequency shifting, and time scaling. These properties help in manipulating and simplifying the Laplace Transform of complex functions. For instance, the Laplace Transform converts differentiation into multiplication by s, the complex frequency variable. This is fundamental in transforming differential equations into algebraic equations, which are typically more straightforward to solve.

Once the algebraic equation is solved, the inverse Laplace Transform is used to convert back to the time domain, often involving partial fraction decomposition to handle the algebraic complexities. Memorization of key transform pairs and the use of transform tables can expedite this process significantly.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are foundational in expressing many physical and engineering problems where change is a central concept, such as in the fields of fluid dynamics, economics, and quantum mechanics.

There are two main types of differential equations: ordinary differential equations (ODEs), which involve functions of a single variable and their derivatives, and partial differential equations (PDEs), involving functions of multiple variables and their partial derivatives. Solutions to differential equations can be complex and often require specialized methods for solving.

The Laplace Transform is an essential tool for solving linear ODEs, especially those with initial value problems. By transforming the differential equations to an algebraic context, the Laplace Transform helps avoid the direct integration of the equation, which can be impractical or impossible for complex equations. After solving the algebraic equation in the Laplace domain, the inverse transform is used to obtain the solution to the original differential equation in the time domain.