A LTI system described by the following differential equation, where x(t) is the input to system and y(t) is output of systemy'(t) + 3y(t) = x(t)When y (0) = 2, the output y (t) for an unit step input is

\(\left( {\frac{1}{4} + \frac{3}{4}{e^{ - 2t}}} \right)u\left( t \right)\)

\(\left( {\frac{1}{3}{e^{ - t}} + \frac{5}{3}{e^{ - 3t}}} \right)u\left( t \right)\)

\(\left( {\frac{1}{4}{e^{ - t}} + \frac{3}{4}{e^{ - 2t}}} \right)u\left( t \right)\)

\(\left( {\frac{1}{3} + \frac{5}{3}{e^{ - 3t}}} \right)u\left( t \right)\)

A LTI system described by the following differential equation, where x

1.	A LTI system described by the following differential equation, where x(t) is the input to system and y(t) is output of systemy'(t) + 3y(t) = x(t)When y (0) = 2, the output y (t) for an unit step input is
A.	\(\left( {\frac{1}{3}{e^{ - t}} + \frac{5}{3}{e^{ - 3t}}} \right)u\left( t \right)\)
B.	\(\left( {\frac{1}{4}{e^{ - t}} + \frac{3}{4}{e^{ - 2t}}} \right)u\left( t \right)\)
C.	\(\left( {\frac{1}{3} + \frac{5}{3}{e^{ - 3t}}} \right)u\left( t \right)\)
D.	\(\left( {\frac{1}{4} + \frac{3}{4}{e^{ - 2t}}} \right)u\left( t \right)\)
Answer» D. \(\left( {\frac{1}{4} + \frac{3}{4}{e^{ - 2t}}} \right)u\left( t \right)\)