Consider the output of an LTI system is YO)=(e-t-e-2t) u{t), to the input signal x(t) = e-3tu(t), determine the system functionH(s), whether or not the system is casual and stable the linear constant- coefficient differential equation of this LTI system.​

Sorry dude but I can't even understand the question

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

it would open into a square depending on the type of envelope i’m not sure about the second part