Let y = y(x) be the solution of the differential equation,

$${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$$

such that y(0) = 0. If $$\sqrt ay(1)$$ = $$\pi \over 32$$ , then the value of 'a' is :

If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as $${{{x^2} - 2y} \over x}$$, then the curve also passes through the point :

Let y = y(x) be the solution of the differential equation, x$${{dy} \over {dx}}$$ + y = x log_e x, (x > 1). If 2y(2) = log_e 4 $$-$$ 1, then y(e) is equal to :

The solution of the differential equation,

$${{dy} \over {dx}}$$ = (x – y)², when y(1) = 1, is :