Consider the differential equation $$\left( {{t^2} - 81} \right){{dy} \over {dt}} + 5ty = \sin \left( t \right)\,\,$$ with $$y\left( 1 \right) = 2\pi .$$ There exists a unique solution for this differential equation when $$t$$ belongs to the interval

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