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Magazine Chapter 69: Problem 5
Determine (a) \(\mathcal{L}^{-1}\left\\{\frac{3}{s^{2}-4 s+13}\right\\}\) (b) \(\mathcal{L}^{-1}\left\\{\frac{2(s+1)}{s^{2}+2 s+10}\right\\}\)
Short Answer
Expert verified
(a) \(3e^{2t}\sin(3t)\), (b) \(2e^{-t}\cos(3t)\)."
Step by step solution
01
Identify the Form of the Inverse Laplace Transform
For problem (a), we need to identify the expression to match it with a known inverse Laplace transform. Notice \[\frac{3}{s^2 - 4s + 13}\] has the form \(\frac{a}{(s-\alpha)^2 + \beta^2}\).
02
Complete the Square for the Denominator
Complete the square for the expression in the denominator:\[s^2 - 4s + 13 = (s - 2)^2 + 9.\]This simplifies our expression to:\[\frac{3}{(s-2)^2 + 3^2}.\]
03
Use the Inverse Laplace Transform Formula
The inverse Laplace transform of \(\frac{a}{(s - \alpha)^2 + \beta^2}\) is \(ae^{\alpha t}\sin(\beta t)\). For \(\alpha = 2, \beta = 3, a = 3\), apply the formula:\[3e^{2t}\sin(3t).\]
04
Identify the Form of the Inverse Laplace Transform
For problem (b), match the expression:\[\frac{2(s+1)}{s^2 + 2s + 10}\] with known forms. This suggests a shifted exponential term along with sine/cosine.
05
Complete the Square for the Denominator
Complete the square:\[s^2 + 2s + 10 = (s + 1)^2 + 3^2.\]This transforms the expression to:\[\frac{2(s+1)}{(s+1)^2 + 3^2}.\]
06
Apply Inverse Laplace Transform Formula
For \(\frac{(s + \alpha)}{(s + \alpha)^2 + \beta^2}\), the inverse Laplace transform is \(e^{-\alpha t}\cos(\beta t)\). For \(\alpha = -1, \beta = 3\), we have:\[2e^{-t}\cos(3t).\]
07
Final Verification
For both parts, double-check the expressions match known inverse Laplace transforms.The inverses correctly reflect: (a) \(3e^{2t}\sin(3t)\) (b) \(2e^{-t}\cos(3t)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace Transform
The Laplace Transform is a powerful mathematical tool used to convert functions from the time domain to the frequency domain. It's especially useful for solving differential equations and analyzing systems. By transforming a problem into an algebraic equation, it simplifies the process of finding solutions.
- The Laplace Transform of a function \( f(t) \) is denoted by \( \mathcal{L}\{f(t)\} = F(s) \).
- It integrates functions from 0 to infinity multiplied by \( e^{-st} \).
- Commonly used for initial value problems.
Completing the square
Completing the square is a method used to transform quadratic expressions into perfect square forms. This technique helps simplify problems involving square terms and is useful when looking for inverse Laplace Transforms.
- For the expression \( s^2 - 4s + 13 \), completing the square transforms it to \( (s-2)^2 + 9 \).
- This step adjusts complex expressions to fit known Laplace Transform forms.
- It involves rewriting a quadratic in the form \( (s - a)^2 + b \).
Sine and Cosine Functions
Sine and cosine functions often appear in inverse Laplace Transform solutions, especially when dealing with expressions that resemble the form \( \frac{a}{(s - \alpha)^2 + \beta^2} \).
- The inverse Laplace of \( \frac{a}{(s - \alpha)^2 + \beta^2} \) is \( ae^{\alpha t}\sin(\beta t) \).
- These functions describe oscillatory behavior, commonly used in physics and engineering to model waves.
Exponential Functions
Exponential functions, such as \( e^{\alpha t} \), frequently occur in inverse Laplace Transforms. They describe growth or decay processes that are common in natural and engineered systems.
- Exponential terms appear as \( e^{\pm \alpha t} \) alongside sine or cosine functions.
- The parameter \( \alpha \) determines the rate and direction of growth or decay.
- These functions aid in modeling continuous change, like population growth or radioactive decay.
