Answer :

Solution :

Given the differential equation is,

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

By applying the Laplace transform,

s2 Y(s) + 4 Y(s) = 6 R(s)

$\Rightarrow \frac{Y\left(s\right)}{R\left(s\right)}=\frac{6}{s^2+4}$

Poles are the roots of the denominator in the transfer function.

⇒ s2 + 4 = 0

⇒ s = ±2j