The differential equation representing the family of curves $${y^2} = 2d(x + \sqrt d )$$, where d is a parameter, is of

Let y(x) be a solution of

$$(1 + {x^2}){{dy} \over {dx}} + 2xy - 4{x^2} = 0$$. Then y(1) is equal to

If $$y = {e^{m{{\sin }^{ - 1}}x}}$$ then $$(1 - {x^2}){{{d^2}y} \over {d{x^2}}} - x{{dy} \over {dx}} - $$ky = 0, where k is equal to

Solution of $${(x + y)^2}{{dy} \over {dx}} = {a^2}$$ ('a' belong a constant) is