Question

Use Theorem 7.1.1 to find ℒ{f(t)}.

 (Write your answer as a function of s.)

f(t) = (t + 1)³

Answer 1

$\text{Note: } f(t)= (t+1)^3 = t^3 +3t^2 +3t +1\\ \text{Recall } \mathcal{L}(t^n) = \frac{n!}{s^{n+1}}\\ \text{Then } \mathcal{L}\{f(t)\} = \mathcal{L}\{t^3 + 3t^2 +3t +1\} = \mathcal{L} \{t^3\} + \mathcal{L} \{3t^2\} + \mathcal{L} \{3t\} +\mathcal{L} \{1\}\\ \mathcal{L} \{f(t)\} = \mathcal{L} \{t^3\} +3\mathcal{L} \{t^2\} +3\mathcal{L} \{t\} +\mathcal{L} \{1\} = \frac{3!}{s^{4}}+ \frac{3\cdot 2!}{s^{3}} + \frac{3}{s^2} + \frac{1}{s} = \frac{6}{s^4} + \frac{6}{s^3} + \frac{3}{s^2} + \frac{1}{s}\\ \therefore \mathcal{L} \{f(t)\} = \frac{6+6s+3s^2+s^3}{s^4}$