The particular solution of the initial value problem given below is $$\,\,{{{d^2}y} \over {d{x^2}}} + 12{{dy} \over {dx}} + 36y = 0\,\,$$ with $$\,y\left( 0 \right) = 3\,\,$$ and $$\,\,{\left. {{{dy} \over {dx}}} \right|_{x = 0}} = - 36\,\,$$

The Solution of the differential equation $$\,{{{d^2}y} \over {d{t^2}}} + 2{{dy} \over {dt}} + y = 0\,\,$$ with $$\,y\left( 0 \right) = {y^1}\left( 0 \right) = 1\,\,$$ is

Consider the differential equation $${{{d^2}x\left( t \right)} \over {d{t^2}}} + 3{{dx\left( t \right)} \over {dt}} + 2x\left( t \right) = 0$$
Given $$x(0) = 20$$ & $$\,x\left( 1 \right) = {{10} \over e},$$ where $$e=2.718,$$

The value of $$x(2)$$ is

With initial values $$\,\,\,y\left( 0 \right) = y'\left( 0 \right) = 1,\,\,\,$$ the solution of the differential equation $$\,\,{{{d^2}y} \over {d{x^2}}} - 4{{dy} \over {dx}} + 4y = 0\,\,$$ at $$x=1$$ is ________.