Question

\(L^{-1}\left\\{\frac{1}{(s+3)}\right\\}=\ldots \ldots\)

Short answer

The inverse Laplace transform is \( e^{-3t} \).

Step-by-step solution

Identify the Laplace Form

We start by identifying the inverse Laplace transformation we need to find, which is given as \( L^{-1}\left\{\frac{1}{s+3}\right\} \).

Recall the Standard Inverse Laplace Transform

Recognize that the standard inverse Laplace transform formula for \( \frac{1}{s-a} \) is \( e^{at} \). In this form, \( a \) represents a constant.

Identify the Parameter 'a'

In the given problem \( \frac{1}{s+3} \), we can equate \( s+3 \) with \( s-a \). Thus, we can see that \( a = -3 \).

Apply the Standard Formula

Using the standard inverse Laplace transform \( L^{-1}\left\{\frac{1}{s-a}\right\} = e^{at} \), substitute \( a = -3 \) to get the inverse Laplace transform as \( e^{-3t} \).

Conclusion

Therefore, the result of the inverse Laplace transform \( L^{-1}\left\{\frac{1}{s+3}\right\} \) is \( e^{-3t} \).

Key concepts

Laplace transformation

Laplace transformation is a powerful technique used to transform a function of time, often a complex differential equation, into a simpler form in the s-domain. It is widely used in engineering and physics to solve ordinary differential equations by converting them into algebraic equations. This makes the process of solving equations more manageable, as algebraic operations are generally simpler than their differential counterparts. The Laplace transformation is given by the formula \( L\{f(t)\} = F(s) \), where \( f(t) \) is a function in the time domain, and \( F(s) \) is its corresponding representation in the frequency domain. By applying this transformation, we can focus on analyzing the frequency components of a signal, which can be crucial in understanding system behaviors in control systems, signal processing, and more.

• Converts differential equations to algebraic equations.
• Contains the variable 's' for easier manipulation of functions.
• Provides a bridge between time-domain and frequency-domain analysis.

Exponential function

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They are closely linked to Laplace transforms, as often the solutions to differential equations involve exponentials. In the solution of the inverse Laplace transformation, the exponential function \( e^{at} \) appears due to its ease of differentiation and integration. This property makes exponential functions essential in modeling growth and decay processes in natural sciences and finance. For instance, exponential decay can describe a decrease in a population or radioactive material over time.

• Has the form \( e^{at} \) where \( a \) is a constant and \( t \) is the variable.
• Often models growth (e.g., populations) or decay (e.g., radioactive decay).
• Is highly significant in the solutions of differential equations.

Mathematical problem solving

Mathematical problem solving in the context of inverse Laplace transformations involves strategic steps to transition from an initial complex form to a recognizable standard formula. The key to successfully solving such a problem is identifying the standard forms and then applying the correct formulas. For example, in our original exercise, solving the inverse Laplace transform \( L^{-1}\left\{\frac{1}{s+3}\right\} \) involves identifying the parameter associated with the standard inverse Laplace formula, \( \frac{1}{s-a} \). This enables the transformation back to the time domain. With practice, students can effectively tackle similar problems using logical reasoning and familiar patterns.

• Identifies the form of the function for transformation.
• Relies on recognizing and applying standard formulas.
• Involves step-by-step logical reasoning.

Standard inverse Laplace formula

The standard inverse Laplace formula is a crucial tool in reversing the Laplace transformation process. This formula helps in converting functions from the s-domain back to the time domain. Specifically, it is defined by \( L^{-1}\left\{\frac{1}{s-a}\right\} = e^{at} \). In practice, you equate a given function to the standard form to identify the parameter \( a \). Once identified, substituting \( a \) into the inverse formula gives you the function in the time domain. For example, if \( a = -3 \) for the function \( \frac{1}{s+3} \), then the inverse is \( e^{-3t} \). Understanding and applying this standard formula is essential to solving problems efficiently in fields such as electrical engineering and applied mathematics.

• Reverses Laplace transforms back to time-domain functions.
• Based on equating to \( \frac{1}{s-a} \) and solving for \( a \).
• Significant in engineering and scientific applications.