Let $y=y(x)$ be the solution of the differential equation $x \sqrt{1-x^2} d y+\left(y \sqrt{1-x^2}-x \cos ^{-1} x\right) d x=0, x \in(0,1), \lim _{x \rightarrow 1^{-}} y(x)=1$. Then $y\left(\frac{1}{2}\right)$ equals :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be such that $f(x y)=f(x) f(y)$, for all $x, y \in \mathbf{R}$ and $f(0) \neq 0$. Let $g:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function such that

$$ x^2 g(x)=\int_1^x\left(\mathrm{t}^2 f(\mathrm{t})-\operatorname{tg}(\mathrm{t})\right) d t $$

Then $g(2)$ is equal to :

Let $f:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function defined as $f(x)=\int_1^x f(\mathrm{t}) \mathrm{dt}+(1-x)\left(\log _{\mathrm{e}} x-1\right)+\mathrm{e}$.

Then the value of $f(f(1))$ is :

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

$$ \frac{d y}{d x}+\left(\frac{6 x^2+\left(3 x^2+2 x^3+4\right) e^{-2 x}}{\left(x^3+2\right)\left(2+e^{-2 x}\right)}\right) y=2+e^{-2 x} $$

$x \in(-1,2)$, satisfying $y(0)=\frac{3}{2}$. If $y(1)=\alpha\left(2+e^{-2}\right)$, then $\alpha$ is equal to :