First, rewrite $ cos(2t) \, $ as a complex exponentional :
- $ cos(2t) =\frac{[e^{j2x}+e^{-j2x}]}{2} $
Then, applying the system to the exponentials gives response $ y(t)\! $:
- $ y(t) = \frac{[te^{-j2x}+te^{j2x}]}{2} $, (because the system is linear, the $ \frac{1}{2} $ factor remains)
Finally, factoring out the $ t\! $ and simplifying back into a sinusoid function yields:
- $ y(t) = tcos(2t)\! $
