The given system equation is

$\dfrac {d^2 y(t)}{dt^2} + 5\dfrac{dy(t)}{dt} + 6y(t)=x(t)$

Taking Laplace transform of the above equation

$[s^2 Y(s) – s y(0) – y’(0)] + 5[s Y(s) – y(0)] + 6Y(s) = X(s)$ ------------ 1

Considering all the initial values zero

Therefore equation 1 becomes

$s^2$ Y(s) + 5s Y(s) + 6Y(s) = X(s)

[$s^2$ + 5s + 6]Y(s) = X(s)

∴$ \frac{Y(s)}{X(s)} = \frac{1}{{s^2}+ 5s + 6}$

∴$ H(s) = \frac{1}{{s^2}+ 5s + 6}$ ------------------ 2

This is the required transfer function of the given system.

Equation 2 can be written as

$ H(s) = \frac{1}{(s+2)(s+3)}$

From the above equation, the value of H(s) will never be zero.

Therefore, the system has no zeros.

From the above equation, the value of H(s) will be infinite for s = -2 and s = -3. Therefore, these are the values of poles.

∴Poles of the system are at s = -2 and s = -3.

Now, x(t) = u(t) , Taking Laplace transform of x(t)

X(s) = $\frac{1}{3}$ -------------------- 3

By convolution theorem,

Y(s) = H(s).X(s)

Substituting the values of H(s) and X(s) from equations 2 and 3

Y(s) = $\frac{1}{s^2 + 5s +6} \frac{1}{s}$

Y(s) = $\frac{1}{s(s+2)(s+3)} $

Expanding above equation in partial fractions

Y(s) = $\frac{k_o}{s} + \frac{k_1}{s+2} + \frac{k_2}{s+3}$ ----------------- 4

Here $k_o$ = s Y(s) | s = 0

= $s \frac{1}{s(s+2)(s+3)}$ | s=0

∴ $k_o = \frac{1}{6}$

$k_1 = (s+2) Y(s)$ | s = -2

= (s+2)$\frac{1}{s(s+2)(s+3)}$ | s = -2

= $\frac{1}{(-2)(-2 + 3)}$

∴ $k_1 = \frac{-1}{2}$

$k_2 = (s+3) \frac{1}{s(s+2)(s+3)}$ | s = -3

= $\frac{1}{(-3)(-3+2)}$

∴ $k_2 = \frac{1}{3}$

Substituting the values of $k_o$, $k_1$ and $k_2$ in equation 4, we get

Y(s) = $ \frac{1}{6} \frac{1}{s} - \frac{1}{2} \frac{1}{s+2} + \frac{1}{3} \frac{1}{s+3} $ ------ ----- 5

Taking inverse Laplace transform of above equation

∴ y(t) = $ \frac{1}{6}u(t) - \frac{1}{2} e^{-2t} u(t) + \frac{1}{3} e^{-3t} u(t) $

This is the response of the system.