The unilateral Laplace transform of $$f(t)$$ is
$$\,{1 \over {{s^2} + s + 1}}.$$ The unilateral Laplace transform of $$t$$ $$f(t)$$ is

Consider the differential equation
$${{{d^2}y\left( t \right)} \over {d{t^2}}} + 2{{dy\left( t \right)} \over {dt}} + y\left( t \right) = \delta \left( t \right)$$
with $$y\left( t \right)\left| {_{t = 0} = - 2} \right.$$ and $${{dy} \over {dt}}\left| {_{t = 0}} \right. = 0.$$

The numerical value of $${{dy} \over {dt}}\left| {_{t = 0}.} \right.$$ is

If $$x = \sqrt { - 1} ,\,\,$$ then the value of $${X^x}$$ is

Given $$f\left( z \right) = {1 \over {z + 1}} - {2 \over {z + 3}}.$$ If $$C$$ is a counterclockwise path in the $$z$$-plane such that
$$\left| {z + 1} \right| = 1,$$ the value of $${1 \over {2\,\pi \,j}}\oint\limits_c {f\left( z \right)dz} $$ is