Question

Find the inverse Laplace transform of each of the following: (a) \(\frac{4}{s^{2}}\) (b) \(-\frac{3}{s}\) (c) \(\frac{5}{s^{3}}\) (d) \(\frac{3}{s^{5}}-\frac{2}{s^{4}}\) (e) \(\frac{1}{3 s^{5}}-\frac{2}{3 s^{2}}\)

Short answer

Question: Determine the inverse Laplace transform of the following functions: (a) \(\frac{4}{s^{2}}\) (b) \(-\frac{3}{s}\) (c) \(\frac{5}{s^{3}}\) (d) \(\frac{3}{s^{5}}-\frac{2}{s^{4}}\) (e) \(\frac{1}{3 s^{5}}-\frac{2}{3 s^{2}}\) Provide your answers in the form of \(L^{-1}\{ F(s) \}=f(t)\) Answer: (a) \(L^{-1}\{\frac{4}{s^{2}}\}=4t\) (b) \(L^{-1}\{-\frac{3}{s}\}=-3\) (c) \(L^{-1}\{\frac{5}{s^{3}}\}=\frac{5t^2}{2}\) (d) \(L^{-1}\{\frac{3}{s^{5}}-\frac{2}{s^{4}}\}=\frac{t^4}{8}-\frac{t^3}{3}\) (e) \(L^{-1}\{\frac{1}{3 s^{5}}-\frac{2}{3 s^{2}}\}=\frac{t^4}{72}-\frac{4}{3}t\)

Step-by-step solution

(a) Inverse Laplace transform of \(\frac{4}{s^{2}}\)

Using the Laplace transform table, we can find that the inverse Laplace transform of \(\frac{1}{s^{2}}\) is \(t\). So for \(\frac{4}{s^{2}}\), the inverse Laplace transform would be \(4t\). Therefore, \(L^{-1}\{\frac{4}{s^{2}}\}=4t\).

(b) Inverse Laplace transform of \(-\frac{3}{s}\)

Using the Laplace transform table, the inverse Laplace transform of \(\frac{1}{s}\) is \(1\). So for \(-\frac{3}{s}\), the inverse Laplace transform would be \(-3\). Therefore, \(L^{-1}\{-\frac{3}{s}\}=-3\).

(c) Inverse Laplace transform of \(\frac{5}{s^{3}}\)

Using the Laplace transform table, the inverse Laplace transform of \(\frac{1}{s^{3}}\) is \(\frac{t^2}{2}\). So for \(\frac{5}{s^{3}}\), the inverse Laplace transform would be \(\frac{5t^2}{2}\). Therefore, \(L^{-1}\{\frac{5}{s^{3}}\}=\frac{5t^2}{2}\).

(d) Inverse Laplace transform of \(\frac{3}{s^{5}}-\frac{2}{s^{4}}\)

Since the inverse Laplace transform is a linear operator, we can find the inverse Laplace transform of each term separately and then combine the results. Using the Laplace transform table, the inverse Laplace transform of \(\frac{1}{s^{5}}\) is \(\frac{t^4}{24}\) and the inverse Laplace transform of \(\frac{1}{s^{4}}\) is \(\frac{t^3}{6}\). So, the inverse Laplace transform of \(\frac{3}{s^{5}}-\frac{2}{s^{4}}\) would be \(L^{-1}\{\frac{3}{s^{5}}\}-L^{-1}\{\frac{2}{s^{4}}\}=\frac{3t^4}{24}-\frac{2t^3}{6}=\frac{t^4}{8}-\frac{t^3}{3}\).

(e) Inverse Laplace transform of \(\frac{1}{3 s^{5}}-\frac{2}{3 s^{2}}\)

Similar to part (d), we can find the inverse Laplace transform of each term separately and then combine the results. Using the Laplace transform table, the inverse Laplace transform of \(\frac{1}{3 s^{5}}\) is \(\frac{t^4}{72}\) and the inverse Laplace transform of \(\frac{2}{3 s^{2}}\) is \(\frac{4}{3}t\). So, the inverse Laplace transform of \(\frac{1}{3 s^{5}}-\frac{2}{3 s^{2}}\) would be \(L^{-1}\{\frac{1}{3 s^{5}}\}-L^{-1}\{\frac{2}{3 s^{2}}\}=\frac{t^4}{72}-\frac{4}{3}t\).

Key concepts

Laplace transform table

The Laplace transform table is an essential tool for engineers and mathematicians working with differential equations. It provides a quick reference guide to find the Laplace and inverse Laplace transforms of common functions. Essentially, it consists of two columns where one lists known functions and the other lists their corresponding Laplace transforms.

Using the table helps simplify the process of finding inverse Laplace transforms and transitioning from the frequency domain back to the time domain. For instance, the table tells us that the inverse Laplace transform of \( \frac{1}{s^2} \) is \( t \), which makes calculating the transform of \( \frac{4}{s^2} \) straightforward since it's simply \( 4t \).

Having a good grasp of how to use this table can significantly speed up your problem-solving process. You can find transformations for basic functions and apply linearity to construct solutions for more complex expressions.

linear operator

The concept of a linear operator is fundamental in understanding how the Laplace transform works, especially when dealing with inverse transforms. When we talk about a linear operator, we mean that it satisfies two main properties: additivity and scalability.

Additivity means that the Laplace transform of a sum of functions equals the sum of their transforms: \( L \{ f(t) + g(t) \} = L \{ f(t) \} + L \{ g(t) \} \). Scalability means applying constants: \( L \{ c \cdot f(t) \} = c \cdot L \{ f(t) \} \), where \( c \) is a constant.

For example, when finding the inverse Laplace transform of a combined expression like \( \frac{3}{s^5} - \frac{2}{s^4} \), we use the linearity of the transform. We individually find the transforms of each part and then combine them to reach the solution in the time domain.

partial fractions decomposition

Partial fractions decomposition is a technique often used in conjunction with the Laplace transform when dealing with rational expressions. It involves breaking down a complex fraction into a sum of simpler fractions, making the inverse Laplace transform easier to find.

This method is particularly useful when the numerator is of a lower degree than the denominator (making it a proper fraction). For instance, consider a function like \( \frac{s}{s^3 + 2s^2 + s} \). Partial fractions decomposition would break this expression into a simpler form such as \( \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2} \), which can then be individually transformed using the Laplace transform table.

This simplification not only facilitates the calculation of the inverse but also enhances understanding of the underlying dynamics in the time domain.

time domain solutions

Time domain solutions refer to the representation of functions in terms of time, typically emerging after applying an inverse Laplace transform. When working with Laplace transforms, we often switch between the time domain and the frequency domain.

Using the inverse Laplace transform, we can revert expressions from the frequency domain back to the time domain, providing solutions that reflect real-world, time-based behavior.

For example, if a frequency domain expression like \( \frac{5}{s^3} \) is transformed, it results in the time domain expression \( \frac{5t^2}{2} \). This gives us a function of time, which can be directly applied to various problems such as those involving electrical circuits, mechanical systems, and other physical processes.

Understanding how to switch between these domains is crucial for analyzing and solving dynamic systems.