Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + 2y = f\left( x \right),$$

where $$f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$$

If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is :

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 :