The solution of the differential equation $2 x^2 y \frac{d y}{d x}=\tan \left(x^2 y^2\right)-2 x y^2$, given $y(1)=\sqrt{\frac{\pi}{2}}$ is

Let $f:(0,1) \rightarrow(0,1)$ be a differentiable function such that $f^{\prime}(x) \neq 0 \forall x \in(0,1)$ and $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$. Suppose for all $x$, $\mathop {\lim }\limits_{t \to x} \frac{\int_0^t \sqrt{1-(f(s))^2} d s-\int_0^x \sqrt{1-(f(s))^2} d s}{f(t)-f(x)}=f(x)$. Then the value of $f\left(\frac{1}{4}\right)$ belongs to

If $x=\int\limits_0^y \frac{1}{\sqrt{1+9 t^2}} d t$ and $\frac{d^2 y}{d x^2}=a y$, then $a$ is equal to

Let $$\mathrm{f}$$ be a differential function with $$\lim _\limits{x \rightarrow \infty} \mathrm{f}(x)=0$$. If $$\mathrm{y}^{\prime}+\mathrm{yf}^{\prime}(x)-\mathrm{f}(x) \mathrm{f}^{\prime}(x)=0$$, $$\lim _\limits{x \rightarrow \infty} y(x)=0$$ then