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 _______.

A curve passes through the point ($$x$$ = 1, $$y$$ = 0) and satisfies the differential equation

$${{dy} \over {dx}} = {{{x^2} + {y^2}} \over {2y}} + {y \over x}$$. The equation that describes the curve 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 $$\,\,{\left. {{{dy} \over {dx}}} \right|_{x = 0}} = - 36\,\,$$