A system is described by the differential equation $$${{{d^2}y} \over {d{t^2}}} + 5{{dy} \over {dt}} + 6y\left( t \right) = x\left( t \right)$$$
Let x(t) be a rectangular pulse given by $$$x\left( t \right) = \left\{ {\matrix{ {1\,\,\,\,\,\,\,\,\,0 \le \,t\, \le 2} \cr {0\,\,\,\,\,otherwise} \cr } } \right.$$$

Assuming that y(0) = 0 $${{dy} \over {dt}} = 0$$ at t = 0, the Laplace transform of y(t) is

If $$F\left( s \right) = L\left[ {f\left( t \right)} \right] = {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}}$$ then the initial and final values of f(t) are respectively

Given f(t) = $${L^{ - 1}}\left[ {{{3s + 1} \over {{s^3} + 4{s^2} + \left( {K - 3} \right)s}}} \right].$$ If $$\matrix{ {Lim\,f\,\left( t \right) = 1,} \cr {t \to \infty } \cr } \,\,$$ then the value of K is

Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of $$\int\limits_0^t {f\left( \tau \right)\,d\tau } $$ is