Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y,$ for $x\ne0.$ If $y(2)=0,$ then $\tan(y(1))$ is equal to

Let the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} y = 2x \exp \left( \frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} \right)$ pass through the origin. Then $y(1)$ is equal to:

Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$,and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then $f(x)$ is

Find a particular solution satisfying the given condition: $\frac{dy}{dx} - 3y \cot x = \sin 2x$; $y = 2$ when $x = \frac{\pi}{2}$.

The solution of the differential equation $x \frac{dy}{dx} = y + x^2$ is

If the general solution of $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$ is $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$, then $f(y) =$