If $$y = {x \over {{{\log }_e}|cx|}}$$ is the solution of the differential equation $${{dy} \over {dx}} = {y \over x} + \phi \left( {{x \over y}} \right)$$, then $$\phi \left( {{x \over y}} \right)$$ is given by

Given $${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$. Changing the independent variable x to z by the substitution $$z = \log \tan {x \over 2}$$, the equation is changed to

The family of curves $$y = {e^{a\sin x}}$$, where 'a' is arbitrary constant, is represented by the differential equation

The solution of

$$\cos y{{dy} \over {dx}} = {e^{x + \sin y}} + {x^2}{e^{\sin y}}$$ is $$f(x) + {e^{ - \sin y}} = C$$ (C is arbitrary real constant) where f(x) is equal to