Use Theorem 7.1.1 to find ℒ{f(t)}.
 (Write your answer as a function of s.)
f(t) = (t + 1)3
"\\text{Note: } f(t)= (t+1)^3 = t^3 +3t^2 +3t +1\\\\\n\\text{Recall } \\mathcal{L}(t^n) = \\frac{n!}{s^{n+1}}\\\\\n\\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\\}\\\\\n\\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}\\\\\n\\therefore \\mathcal{L} \\{f(t)\\} = \\frac{6+6s+3s^2+s^3}{s^4}"
Comments
Leave a comment