Part B: The basics of linearity
$ x_1(t) = e^{2jt} \rightarrow linear-system \rightarrow y_1(t) = te^{-2jt} $
$ x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} $
The input, $ cos(2t) = \frac{1}{2}(e^{j2t} + e^{-j2t}) $
From the properties of a linear system $ ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t) $, where $ a = b = \frac{1}{2} $
The response to cos(2t) is:
$ y(t) = \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $
$ y(t) = t[\frac{1}{2}(e^{-2jt}+e^{2jt})] $
$ y(t) = tcos(2t) $
Therefore, $ x(t) = cos(2t) \rightarrow linear-system \rightarrow y(t) = tcos(2t) $
