Introduction and Overview

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The Laplace transform is a fundamental tool in applied mathematics, engineering, and physics, used to solve differential equations, analyze linear time-invariant systems, and perform signal processing. This article delves into the detailed process of evaluating the Laplace transform of te^{3t}sin(2t).

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Step-by-Step Evaluation of the Laplace Transform

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1. Basic Laplace Transform Fact

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Let's start with a basic Laplace transform fact that is crucial for this problem:

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[mathcal{L}{sin(2t)} frac{2}{s^2 - 2^2}]

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This formula suggests that the Laplace transform of sin(2t) is 2/(s2 - 4).

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2. Applying the Shifting Theorem

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The shifting theorem is an important property of Laplace transforms. It states that:

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[mathcal{L}{e^{3t}f(t)} F(s-3)]r where F(s) is the Laplace transform of f(t).

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To apply the shifting theorem to our function, we need to find the Laplace transform of e3tsin(2t):

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[mathcal{L}{e^{3t}sin(2t)} frac{2}{(s-3)^2 - 2^2}]

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This simplifies to:

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[mathcal{L}{e^{3t}sin(2t)} frac{2}{(s-3)^2 - 4}]

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3. Multiplication by (t)

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The next step is to apply the "multiplication by (t)" rule in Laplace transforms. This rule states that:

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[mathcal{L}{t f(t)} -frac{d}{ds} mathcal{L}{f(t)}]

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Applying this rule to our function:

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[mathcal{L}{t e^{3t}sin(2t)} -frac{d}{ds} left( frac{2}{(s-3)^2 - 4} right)]

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4. Differentiation and Simplification

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To find the derivative, let's use the quotient rule.

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The derivative of F(s) 2/(s2 - 6s 5) is:

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[F'(s) -2 cdot frac{(2s-6)}{(s-3)^2 - 4)^2}]

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Substituting back, we get:

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[mathcal{L}{t e^{3t}sin(2t)} frac{4s-12}{(s-3)^2 - 4)^2}]

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This simplifies further to:

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[mathcal{L}{t e^{3t}sin(2t)} frac{4s-12}{(s-3)^2 - 16}]

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This is the final form of the Laplace transform we were looking for.

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Conclusion

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This step-by-step guide demonstrates how to evaluate the Laplace transform of te3tsin(2t) using basic transformations and the shifting theorem. Understanding these steps and properties is crucial for solving similar problems in engineering and applied mathematics.

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