Solution:

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

Applying Laplace transform of the given equation,

$\begin{gathered} S^2 Y(S)+5 S Y(S)+6 Y(S)=X(S) \\\\ Y(S)\left(S^2+5 S+6\right)=X(S)\\ \end{gathered}\\ $

$\text { Transfer function } H(S)=\frac{Y(S)}{X(S)}=\frac{1}{\left(S^2+5 S+6\right)}\\ $

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

$\begin{gathered}\\ H(S)=\frac{1}{(S+3)(S+2)}=\frac{A}{S+3}+\frac{B}{S+2} \\\\ 1=A(S+2)+B(S+3)\\ \end{gathered}\\ $

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

Applying Inverse Laplace transform,

$h(t)=-e^{-3 t} u(t)+e^{-2 t} u(t)\\ $