For the first term, 3t e^{2t}, we have:
$$
\mathcal{L}\{3t e^{2t}\} = \int_0^\infty 3t e^{2t} e^{-st} dt
$$
Let's substitute u = s - 2, then s = u + 2:
$$
\mathcal{L}\{3t e^{2t}\} = \int_0^\infty 3t e^{-(u+2)t} e^{2t} dt = \int_0^\infty 3t e^{-ut} dt
$$
Now,
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