If a function y(x) is described by the initial value problem, \(\frac{{{d^2}y}}{{d{x^2}}} + 5\frac{{dy}}{{dx}} + 6y = 0\), with initial conditions y(0) = 2, and \({\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = 0,\) then the value of y at x = 1 is ________. (Round off to 2 decimal places)