Determine the Laplace transform of the following function:
$$ f(t) = te^{-t}\sin(2t) $$ Recall the definition of the Frequency Differentiation Property. This is also property #30 in our Laplace Transform table. $$ \mathcal{L}[t^nf_1(t)] = F(s) = (-1)^n \frac{d^nF_1(s)}{ds^n} $$ We start by determing the Laplace Transform of f1(t). This is the expression f(t) without the "t" to the nth power part of the expression
$$ \begin{align} \mathcal{L}[f_1(t)] & = F_1(s) \\ & = \mathcal{L}[e^{-t}\sin(2t)] \qquad,(Eqn\;1)\\ \end{align} $$ When examining equation #1, we recognize that we can utilize the Frequency Shift Property as well as property #19 in our Laplace Transform table. Continuing from equation #1 we have: $$ \begin{align} F_1(s) & = \frac{2}{(s+1)^2 + 2^2} \\ F_1(s) & = \frac{2}{(s+1)^2 + 4} \qquad,(Eqn\;2)\\ \end{align} $$ We now plug equation #2 into our definition for the Frequency Differentiation Property in order to determine F(s): $$ \begin{align} \mathcal{L}[te^{-t}\sin(2t)] & = F(s) \\ & = (-1)^{-1}\frac{d}{ds}\Big[ \frac{2}{(s+1)^2 + 4} \Big] \\ & = -1 \Big[ \frac{0[(s+1)^2 + 4]-2(2)(s+1)}{[(s+1)^2 + 4]^2} \Big] \\ & = -1 \Big[ \frac{-4s-4}{[(s+1)^2 + 4]^2} \Big] \end{align} $$ Finishing up, we finally arrive at our final solution:
$$ F(s) = \frac{4s+4}{[(s+1)^2 + 4]^2} $$
Continue on to Properties of the Laplace Transform example problem #3...