The position of a particle y(t) is described by the differential equation :

$${{{d^2}y} \over {d{t^2}}} = - {{dy} \over {dt}} - {{5y} \over 4}$$.

The initial conditions are y(0) = 1 and $${\left. {{{dy} \over {dt}}} \right|_{t = 0}}$$ = 0.

The position (accurate to two decimal places) of the particle at t = $$\pi $$ is _______.

Which one of the following is the general solution of the first order differential equation $${{dy} \over {dx}} = {\left( {x + y - 1} \right)^2}$$ , where $$x,$$ $$y$$ are real ?

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 $$\,\,{\left. {{{dy} \over {dx}}} \right|_{x = 0}} = - 36\,\,$$

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