Math Problem Statement
Solution
To find the Laplace transform of the given piecewise function:
\begin{cases} 0, & t < 2 \\ t^2 - 4t + 9, & t \geq 2 \end{cases}$$ ### Step-by-Step Approach: Since the function $$f(t)$$ is zero before $$t = 2$$, we need to shift and handle the function using the **Heaviside (unit step) function**. We can rewrite the given function in terms of the Heaviside function $$u(t - 2)$$, which activates the non-zero part of the function at $$t = 2$$: $$f(t) = \left(t^2 - 4t + 9\right) u(t - 2).$$ Now, we use the **time-shifting property of the Laplace transform**: ### Time-Shifting Property: If $$f(t) = g(t - a) u(t - a)$$, then: $$\mathcal{L}\{f(t)\} = e^{-as} \mathcal{L}\{g(t)\}.$$ In our case, $$a = 2$$ and $$g(t) = t^2 - 4t + 9$$. However, since $$t$$ shifts to $$t - 2$$ under the Heaviside function, we must first replace $$t$$ with $$(t + 2)$$ in the function $$t^2 - 4t + 9$$. ### Step 1: Replace $$t$$ with $$t + 2$$ The expression becomes: $$(t + 2)^2 - 4(t + 2) + 9.$$ Simplifying: $$t^2 + 4t + 4 - 4t - 8 + 9 = t^2 + 5.$$ Thus, $$g(t) = t^2 + 5$$. ### Step 2: Find the Laplace Transform of $$t^2 + 5$$ Using standard Laplace transform rules: $$\mathcal{L}\{t^2\} = \frac{2}{s^3}, \quad \mathcal{L}\{5\} = \frac{5}{s}.$$ Thus: $$\mathcal{L}\{t^2 + 5\} = \frac{2}{s^3} + \frac{5}{s}.$$ ### Step 3: Apply the Time-Shift Property Using the time-shift property: $$\mathcal{L}\{(t^2 - 4t + 9) u(t - 2)\} = e^{-2s} \left(\frac{2}{s^3} + \frac{5}{s}\right).$$ ### Final Answer: The Laplace transform of the given function is: $$\boxed{e^{-2s} \left(\frac{2}{s^3} + \frac{5}{s}\right)}.$$ Let me know if you want further clarification or details on any step! --- Here are 5 related questions: 1. How do you derive the Laplace transform of a polynomial function? 2. What is the role of the Heaviside function in Laplace transforms? 3. How does the time-shifting property of Laplace transforms work? 4. How can Laplace transforms be applied to solve differential equations? 5. What are common mistakes when working with piecewise functions in Laplace transforms? **Tip:** Always remember to adjust the limits and variables properly when working with the Heaviside function in Laplace transforms.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Laplace Transform
Piecewise Functions
Heaviside Step Function
Time Shifting Property
Formulas
Laplace Transform of f(t): L{f(t)} = ∫(0, ∞) f(t) e^(-st) dt
Time Shifting Property: L{f(t - a) u(t - a)} = e^(-as) L{f(t)}
Laplace Transform of Polynomial Functions: L{t^n} = n!/s^(n+1)
Theorems
Time-Shifting Property
Laplace Transform of Heaviside Step Function
Suitable Grade Level
Undergraduate
Related Recommendation
Laplace Transform of Piecewise Function with Heaviside Step Function
Laplace Transform of Piecewise Function f(t) = 0 for t < 4 and t^2 - 8t + 22 for t ≥ 4
Laplace Transform of a Piecewise Function Using Heaviside Step Functions
Laplace Transform of Piecewise Function t^2 for 0 <= t < 2, 1 - t - 3t^2 for t >= 2
Laplace Transform of a Shifted Quadratic Function Using Heaviside Function