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 analyze its differential equation:
$\frac{d^{2}y\left(t\right)}{d{t}^2}+4y\left(t\right)=6r\left(t\right)$

First, we apply the Laplace transform to the differential equation, assuming zero initial conditions. Recall that the Laplace transform of the second derivative of a function $y\left(t\right)$ is $s^{2}Y\left(s\right)$, and the Laplace transform of $y\left(t\right)$ is $Y\left(s\right)$. Applying this transformation gives:
$s^{2}Y\left(s\right)+4Y\left(s\right)=6R\left(s\right)$

Factoring out $Y\left(s\right)$ on the left side of the equation yields:
$(s^2+4)Y\left(s\right)=6R\left(s\right)$

The transfer function $H\left(s\right)$ of the system is the ratio of the output Laplace transform $Y\left(s\right)$ to the input Laplace transform $R\left(s\right)$:
$H\left(s\right)=\frac{Y\left(s\right)}{R\left(s\right)}=\frac{6}{s^2+4}$

The poles of the system are the roots of the characteristic equation, which is obtained by setting the denominator polynomial of the transfer function to zero:
$s^2+4=0$

Solving for $s$:
$s^2=-4$
Taking the square root of both sides gives:
$s=\pm \sqrt{-4}$
Using the imaginary unit $j=\sqrt{-1}$, we find:
$s=\pm 2j$

Thus, the poles of the system are located at $+2j$ and $-2j$.