$A$ function $y=f(x)$ satisfies $f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$ with the condition $f(0) = 0$. Then $f(\frac{\pi}{2})$ is equal to

Let a curve $y = y(x)$ pass through the point $(3,3)$ and the area of the region under this curve,above the $x$-axis and between the abscissae $3$ and $x (>3)$ be $\left(\frac{y}{x}\right)^{3}$. If this curve also passes through the point $(\alpha, 6\sqrt{10})$ in the first quadrant,then $\alpha$ is equal to $........$

Let $y = y(x)$ be the solution of the differential equation $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$,where $x \in (0, 1)$ and $\lim_{x\to 1^-} y(x) = 1$. Then $y\left(\frac{1}{2}\right)$ equals:

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

Find the integrating factor of the differential equation $(1+x^{2}) dt = (\tan^{-1} x - t) dx$.

The differential equation $\frac{dy}{dx} = \frac{x+y}{1+x^2}$ is a . . . . . . differential equation.