Question

In Exercises \(7-18\) express the given function \(f\) in terms of unit step functions and use Theorem 8.4 .1 to find \(\mathcal{L}(f) .\) Where indicated by, graph \(f\). $$ f(t)=\left\\{\begin{array}{cc} 0, & 0 \leq t<2 \\ t^{2}+3 t, & t \geq 2 \end{array}\right. $$

Short answer

Question: Find the Laplace transform of the given function \(f(t)\). Answer: The Laplace transform of the given function \(f(t)\) is \(\mathcal{L}(f) = e^{-2s}\left(\frac{2}{s^3} + \frac{3}{s^2}\right)\).

Step-by-step solution

Express the function in terms of unit step functions

First, let's express the given function in terms of unit step functions. The unit step function is denoted by \(u(t - a)\), where \(a\) is the time at which the function "steps" from 0 to 1. We can express the function \(f(t)\) as a sum of two functions: the first one is equal to \(t^2 + 3t\) when \(t \geq 2\), and the second one is equal to 0 when \(t<2\). Therefore, we can write: $$ f(t) = (t^2 + 3t)u(t - 2) $$

Apply Theorem 8.4.1

Now that we have the function \(f(t)\) in terms of unit step functions, we can apply the theorem 8.4.1 to find the Laplace transform of \(f(t)\). The theorem states: $$ \mathcal{L}\{f(t - a)u(t - a)\} = e^{-as}F(s) $$ where \(F(s) = \mathcal{L}\{f(t)\}\). Here, we have \(a=2\), so we can write: $$ \mathcal{L}\{(t^2 + 3t)u(t - 2)\} = e^{-2s}\mathcal{L}\{t^2 + 3t\} $$

Compute the Laplace transform of the polynomial function

Next, we need to find the Laplace transform of the polynomial function \(t^2+3t\). Recall the formula for finding the Laplace transform of a function of the form \(t^n\): $$ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} $$ Now, we can find the Laplace transform of \(t^2+3t\): $$ \begin{aligned} \mathcal{L}\{t^2 + 3t\} &= \mathcal{L}\{t^2\} + 3\mathcal{L}\{t\} \\ &= \frac{2!}{s^{2+1}} + 3\frac{1!}{s^{1+1}} \\ &= \frac{2}{s^3} + \frac{3}{s^2} \end{aligned} $$

Final expression for the Laplace transform of \(f(t)\)

Now that we have found the Laplace transform of the polynomial function, we can substitute it back into the expression we obtained using theorem 8.4.1: $$ \mathcal{L}\{(t^2 + 3t)u(t - 2)\} = e^{-2s}\left(\frac{2}{s^3} + \frac{3}{s^2}\right) $$ Therefore, the Laplace transform of the given function \(f(t)\) is: $$ \mathcal{L}(f) = e^{-2s}\left(\frac{2}{s^3} + \frac{3}{s^2}\right) $$