The solution curve of the differential equation,

(1 + e^-x)(1 + y²)$${{dy} \over {dx}}$$ = y²,

which passes through the point (0, 1), is :

If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x²dy= (2xy + y²)dx, then $$f\left( {{1 \over 2}} \right)$$ is equal to :

Let y = y(x) be the solution of the differential equation,
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.
If y($$\pi $$) = a and $${{dy} \over {dx}}$$ at x = $$\pi $$ is b, then the ordered pair (a, b) is equal to :

If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :