We are told that a system is linear and given inputs
$ \,x_1(t)=e^{2jt}\, $ yields $ \,y_1(t)=te^{-2jt}\, $
$ \,x_2(t)=e^{-2jt}\, $ yields $ \,y_2(t)=te^{2jt}\, $
- initial given statements copied from Jeff Kubascik
Since,
$ \cos(2t) = \frac {e^{i2t} + e^{-i2t}} {2} $
from the given input we know:
y(t) = t*x(-t)
If we multiply both t's in the equivalent cos(2t) by -1, then they simply switch places leaving the function unchanged. Next, multiply by t and you are given your result.
y(t) = t*cos(2t)
