Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation:

$\frac{d^{2}y\left(t\right)}{d{t}^2}+4y\left(t\right)=6r\left(t\right)$

The poles of this system are at

The correct option is +2j, -2j.

To find the poles of the given linear time-invariant (LTI) system, we first need to determine the system's transfer function by taking the Laplace transform of the governing differential equation, assuming zero initial conditions.

The given differential equation is:

$\frac{d^{2}y(t)}{d{t}^2}+4y(t)=6r(t)$

Taking the Laplace transform on both sides (assuming zero initial conditions):

$s^{2}Y(s)+4Y(s)=6R(s)$

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

$(s^2+4)Y(s)=6R(s)$

The transfer function $H(s)=\frac{Y(s)}{R(s)}$ is given by:

$H(s)=\frac{6}{s^2+4}$

The poles of the system are the roots of the characteristic equation (i.e., setting the denominator of the transfer function to zero):

$s^2+4=0$

Solving for $s$:

$s^2=-4$

$s=\pm \sqrt{-4}=\pm 2j$

Therefore, the poles of the system are located at +2j, -2j.