Math Problem Statement
Find the Laplace transform of f(t) = {0, t < 6; (t - 6)^5, t >= 6}.
Solution
To find the Laplace transform of , we observe that is defined as follows:
0, & t < 6 \\ (t - 6)^5, & t \geq 6 \end{cases}$$ This function can be expressed using the unit step function $$u(t - 6)$$ as: $$f(t) = (t - 6)^5 \cdot u(t - 6)$$ The Laplace transform of $$f(t) = (t - 6)^5 u(t - 6)$$ can be found using the **time-shifting property** of the Laplace transform, which states: $$\mathcal{L} \{ f(t - a) u(t - a) \} = e^{-as} \mathcal{L} \{ f(t) \}$$ In this case, $$f(t - 6) = (t - 6)^5$$ and $$a = 6$$. We first find the Laplace transform of $$t^5$$: $$\mathcal{L} \{ t^5 \} = \frac{5!}{s^{6}} = \frac{120}{s^{6}}$$ Applying the time-shifting property, we get: $$\mathcal{L} \{ (t - 6)^5 u(t - 6) \} = e^{-6s} \cdot \frac{120}{s^6}$$ Thus, $$F(s) = \frac{120 e^{-6s}}{s^6}$$ Would you like a deeper explanation of any step, or have additional questions? --- Here are 5 related questions that expand on this information: 1. How does the time-shifting property work for functions involving the unit step function? 2. What is the significance of the unit step function in Laplace transforms? 3. How would the Laplace transform change if $$f(t) = (t - a)^n u(t - a)$$ for any integer $$n$$? 4. Can we use a similar approach if $$f(t)$$ were defined piecewise with more intervals? 5. What are other applications of the Laplace transform in solving differential equations? **Tip:** The time-shifting property is especially useful when dealing with delayed or piecewise-defined functions in Laplace transformations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Piecewise Functions
Time Shifting
Unit Step Function
Formulas
Laplace Transform of t^n: L{t^n} = n! / s^(n+1)
Time Shifting Property: L{f(t - a) u(t - a)} = e^(-as) L{f(t)}
Theorems
Time Shifting Property of Laplace Transform
Suitable Grade Level
Undergraduate
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