Let a causal LTI system be governed by the following differential equation y(t) + 1/4 dy/dt = 2x(t), where x(t) and y(t) are the input and output respectively. Its impulse response is

The correct option is 8e^-4tu(t).

To find the impulse response of the given causal linear time-invariant (LTI) system, we start with the governing differential equation:

$$y(t)+\frac{1}{4}\frac{dy(t)}{dt}=2x(t)$$

where $x(t)$ is the input and $y(t)$ is the output. To make calculations simpler, we can multiply the entire equation by 4 to obtain:

$$\frac{dy(t)}{dt}+4y(t)=8x(t)$$

To find the transfer function of the system, we apply the Laplace transform to both sides of the differential equation, assuming zero initial conditions:

$$sY(s)+4Y(s)=8X(s)$$

Factoring out $Y(s)$ on the left side gives:

$$(s+4)Y(s)=8X(s)$$

The transfer function $H(s)$ is defined as the ratio of the output Laplace transform to the input Laplace transform:

$$H(s)=\frac{Y(s)}{X(s)}=\frac{8}{s+4}$$

The impulse response $h(t)$ is the inverse Laplace transform of the transfer function $H(s)$. Using the standard transform pair $e^{-at}u(t)\iff \frac{1}{s+a}$ for a causal system (where the region of convergence is Re{s} > -4), we get:

$$h(t)={\mathcal{L}}^{-1}\{\frac{8}{s+4}\}=8e^{-4t}u(t)$$

This matches the correct option.