If 2x = y$${^{{1 \over 5}}}$$ + y$${^{ - {1 \over 5}}}$$ and

(x² $$-$$ 1) $${{{d^2}y} \over {d{x^2}}}$$ + $$\lambda $$x $${{dy} \over {dx}}$$ + ky = 0,

then $$\lambda $$ + k is equal to :

The curve satisfying the differential equation, ydx $$-$$(x + 3y²)dy = 0 and passing through the point (1, 1), also passes through the point :

If $$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$$ and y(0) = 1,

then $$y\left( {{\pi \over 2}} \right)$$ is equal to :

The solution of the differential equation

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :