Let $$y(x)$$ be the solution of the differential equation $$\,\,{{{d^2}y} \over {d{x^2}}} - 4{{dy} \over {dx}} + 4y = 0\,\,$$ with initial conditions $$y(0)=0$$ and $$\,\,{\left. {{{dy} \over {dx}}} \right|_{x = 0}} = 1.\,\,$$ Then the value of $$y(1)$$ is __________.

A function $$y(t),$$ such that $$y(0)=1$$ and $$\,y\left( 1 \right) = 3{e^{ - 1}},\,\,$$ is a solution of the differential equation $$\,\,{{{d^2}y} \over {d{t^2}}} + 2{{dy} \over {dt}} + y = 0\,\,$$ Then $$y(2)$$ is

A solution of the ordinary differential equation $$\,\,{{{d^2}y} \over {d{t^2}}} + 5{{dy} \over {dt}} + 6y = 0\,\,$$ is such that $$y(0)=2$$ and $$y(1)=$$ $$ - \left( {{{1 - 3e} \over {{e^3}}}} \right).$$ The value of $${{dy} \over {dt}}\left( 0 \right)$$ is

A differential equation $$\,\,{{di} \over {dt}} - 0.21 = 0\,\,$$ is applicable over $$\,\, - 10 < t < 10.\,\,$$ If $$i(4)=10,$$ then $$i(-5)$$ is