prove or desprove lenearity..

[latex]

y(t)=e^{-t}x(t)
[/latex]

i need to prove lenearity by constant
and by addition

[latex]

y(t_1)=e^{-t_1}x(t_1)
[/latex]
[latex]

y(t_2)=e^{-t_2}x(t_2)
[/latex]
[latex]

y(t_1)+y(t_2)=e^{-t_1}x(t_1)+e^{-t_2}x(t+2)
[/latex]

[latex]y(t_1+t_2)=e^{-t_1-t_2}x(t_1+t_2)[/latex]

i dont know if [latex]y(t_1+t_2)= y(t_1)+y(t_2)[/latex]
?

transgalactic said:

[latex]

y(t)=e^{-t}x(t)
[/latex]

i need to prove lenearity by constant
and by addition

[latex]

y(t_1)=e^{-t_1}x(t_1)
[/latex]
[latex]

y(t_2)=e^{-t_2}x(t_2)
[/latex]
[latex]

y(t_1)+y(t_2)=e^{-t_1}x(t_1)+e^{-t_2}x(t+2)
[/latex]

[latex]y(t_1+t_2)=e^{-t_1-t_2}x(t_1+t_2)[/latex]

i dont know if [latex]y(t_1+t_2)= y(t_1)+y(t_2)[/latex]
?

Click to expand...

I think you have to multiply the variable t1 or t2 by a constant. so the RHS has eg. 2*t1 in place of t1, and if the result is equal to 2*y then it is linear.

by addition, I am not too sure and wouldnt want to lead you in the wrong direction, but I think you need to prove that [latex]y(t_1+t_2)= y(t_1)+y(t_2)[/latex], which again I would suggest to do by substitution.