Question

Find the inverse Laplace transform of (a) \(\frac{1}{2(s+3)}\) (b) \(\frac{1}{2 s+3}\) (c) \(\frac{3 s}{s^{2}+1}\) (d) \(\frac{-6}{s^{2}+9}\) (e) \(\frac{s+2}{s^{2}+4}\)

Short answer

Question: Find the inverse Laplace transforms of the following functions: (a) \(\frac{1}{2(s+3)}\) (b) \(\frac{1}{2s+3}\) (c) \(\frac{3s}{s^2+1}\) (d) \(\frac{-6}{s^2+9}\) (e) \(\frac{s+2}{s^2+4}\) Answer: (a) \(\frac{1}{2}e^{-3t}\) (b) \(\frac{1}{3}e^{-\frac{3}{2}t}\) (c) \(3\cos{t}\) (d) \(-\frac{2}{3}\sin{3t}\) (e) \(\frac{1}{4} \cos{2t}+\frac{1}{4}\sin{2t}\)

Step-by-step solution

(a) Find the inverse Laplace transform of \(\frac{1}{2(s+3)}\)

To find the inverse Laplace transform of \(\frac{1}{2(s+3)}\), we can use the Laplace transform table. The given function is the Laplace transform of the exponential function \(e^{-3t}\) scaled by \(\frac{1}{2}\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{1}{2(s+3)}\right)=\frac{1}{2}e^{-3t}$$

(b) Find the inverse Laplace transform of \(\frac{1}{2s+3}\)

To find the inverse Laplace transform of \(\frac{1}{2s+3}\), we can rewrite the function as: $$\frac{1}{2s+3} = \frac{1/3}{s+\frac{3}{2}}$$ Now, we recognize that this function is similar to the Laplace transform of the exponential function \(e^{-\frac{3}{2}t}\) scaled by \(\frac{1}{3}\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{1}{2s+3}\right)=\frac{1}{3}e^{-\frac{3}{2}t}$$

(c) Find the inverse Laplace transform of \(\frac{3s}{s^2+1}\)

To find the inverse Laplace transform of \(\frac{3s}{s^2+1}\), we compare this function with the Laplace transform table and find that this function is the Laplace transform of the cosine function scaled by \(3\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{3s}{s^2+1}\right)=3\cos{t}$$

(d) Find the inverse Laplace transform of \(\frac{-6}{s^2+9}\)

To find the inverse Laplace transform of \(\frac{-6}{s^2+9}\), we can rewrite the function as: $$\frac{-6}{s^2+9} = -\frac{2}{3}\frac{3}{s^2+3^2}$$ We recognize that this function is similar to the Laplace transform of the sine function scaled by \(-\frac{2}{3}\) and where \(a=3\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{-6}{s^2+9}\right)=-\frac{2}{3}\sin{3t}$$

(e) Find the inverse Laplace transform of \(\frac{s+2}{s^2+4}\)

To find the inverse Laplace transform of \(\frac{s+2}{s^2+4}\), we first rewrite it as a sum: $$\frac{s+2}{s^2+4} = \frac{1}{4}\frac{4s}{s^2+4^2}+\left(\frac{1}{2}-\frac{1}{4}\right)\frac{1}{s^2+4^2}$$ Now, we identify both functions as Laplace transforms of the cosine function and sine function, respectively, and find the inverse Laplace transforms: $$L^{-1}\left(\frac{s+2}{s^2+4}\right)=\frac{1}{4} \cos{2t}+\frac{1}{4}\sin{2t}$$

Key concepts

Inverse Laplace Transform

The inverse Laplace transform is a mathematical technique used to determine the original function from its Laplace-transformed version. Imagine you've got a puzzle-solving tool that lets you go back from a coded message to its original phrase. It's really handy for solving equations and understanding systems in engineering and science.

To understand the inverse Laplace transform better, think of it as the reverse process of the Laplace transform, which is often used to convert functions from the time domain to the frequency domain. The beauty of the inverse Laplace is its ability to retrieve the time-domain function, allowing us to solve complex differential equations with more ease.

For instance, when taking the inverse Laplace transform of \(\frac{1}{2(s+3)}\), you use knowledge from a Laplace table that reveals this corresponds to the scaled exponential function \(\frac{1}{2}e^{-3t}\). Hence, you apply this process for many algebraic expressions in terms of \(s\), which are simply waiting to be decoded back into useful time functions.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are crucial for modeling how things change over time, like how populations grow or how heat transfers. These equations call for methods like the Laplace transform to be solved more effectively.

Handling differential equations is often about maneuvering through many curves and slopes to find out how a system behaves. By using tools like the Laplace transform, we can simplify these equations. The Laplace transform turns differential equations into algebraic equations, making them easier to solve.

In the solution exercise, taking the inverse Laplace transform skillfully helps translate the expressions into simple functions involving time, like sine and cosine. It directly provides the solutions needed to understand the initial state described by the differential equation. Transform methods enable this transition from complex workings into more direct conclusions.

Transforms and Applications

Transforms are mathematical techniques that convert one form of function representation into another form. In particular, the Laplace transform is a powerful method used to analyze linear time-invariant systems, typical in engineering.

The value of transforms lies in their ability to simplify complex operations. They are applied widely in physics and engineering, especially in analyzing circuits, control systems, and signal processing.

• Transforms help convert convolution operations into mere multiplication, making them much easier to handle.
• They simplify the process of solving linear differential equations, reducing them to algebraic tasks.
• Many real-world phenomena modeled by differential equations become more tractable.

The application seen in the exercise shows the elegance of using transforms to analyze and solve expression forms. By moving through the inverse Laplace, expressions are indeed brought back from their transformed state, yielding practical insights into how systems react and behave over time.