The correct answer is $\boxed{{1 \over 5}{e^{3t}}u\left( { – t} \right) + {1 \over 5}{e^{ – 2t}}u\left( { – t} \right)}$.
The impulse response of an LTI system is the response of the system to a unit impulse input. The differential equation $y”(t) – y'(t) – 6y(t) = x(t)$ can be solved using the method of undetermined coefficients. The general solution to this equation is $y(t) = Ae^{3t} + Be^{-2t}$, where $A$ and $B$ are constants to be determined.
To determine $A$ and $B$, we need to know the initial conditions of the system. In this case, we are told that the system is neither causal nor stable. This means that the system does not have a zero initial condition, i.e., $y(0) \neq 0$.
The initial conditions of the system can be found by taking the derivatives of the general solution and evaluating them at $t = 0$. The first derivative of $y(t)$ is $y'(t) = 3Ae^{3t} – 2Be^{-2t}$. The second derivative of $y(t)$ is $y”(t) = 9Ae^{3t} + 4Be^{-2t}$.
Evaluating these derivatives at $t = 0$, we get $y'(0) = 3A$ and $y”(0) = 9A$. We are also given that $x(t) = u(t)$, where $u(t)$ is the unit step function. The unit step function is defined as $u(t) = 0$ for $t < 0$ and $u(t) = 1$ for $t \geq 0$.
The response of the system to the unit step input is $y(t) = Ae^{3t} + Be^{-2t}$. To find $A$ and $B$, we need to substitute this expression into the differential equation $y”(t) – y'(t) – 6y(t) = x(t)$. This gives us the following equation:
$$9Ae^{3t} – 2Be^{-2t} – 3Ae^{3t} + 2Be^{-2t} – 6(Ae^{3t} + Be^{-2t}) = 1$$
Simplifying this equation, we get $-3A – 4B = 1$. Solving for $A$ and $B$, we get $A = {1 \over 5}$ and $B = {1 \over 5}$.
Therefore, the impulse response of the system is $h(t) = {1 \over 5}{e^{3t}}u\left( { – t} \right) + {1 \over 5}{e^{ – 2t}}u\left( { – t} \right)$.
The other options are incorrect because they do not satisfy the differential equation $y”(t) – y'(t) – 6y(t) = x(t)$.