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).

We are given the differential equation of a causal Linear Time-Invariant (LTI) system:

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

where y(t) is the output and x(t) is the input.

To find the impulse response, we can take the Laplace transform on both sides of the differential equation, assuming initial conditions to be zero:

$Y\left(s\right)+\frac{1}{4}sY\left(s\right)=2X\left(s\right)$

Now, factor out Y(s) on the left side of the equation:

$Y\left(s\right)(1+\frac{1}{4}s)=2X\left(s\right)$

The transfer function H(s) is defined as the ratio of Y(s) to X(s):

$H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{2}{1+\frac{1}{4}s}$

Multiply the numerator and the denominator by 4 to convert it to standard form:

$H\left(s\right)=\frac{8}{s+4}$

To find the impulse response h(t), we take the inverse Laplace transform of H(s). Using the standard Laplace transform pair for causal systems:

$e^{-at}u\left(t\right)\iff \frac{1}{s+a}$

where u(t) is the unit step function and a = 4. Applying this pair and scaling by 8, we get:

$h\left(t\right)=8e^{-4t}u\left(t\right)$