If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to :

If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :

If $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$, x > 0, $$\phi$$ > 0, and y(1) = $$-$$1, then $$\phi \left( {{{{y^2}} \over 4}} \right)$$ is equal to :