The solution of $x^2 + y^2 \frac{dy}{dx} = 4$ is

If the solution of $\frac{dy}{dx} = (3x + y + 4)^2$ is $\frac{1}{\sqrt{3}} \tan^{-1}(f(x, y)) - x = k$,then $f(1, 2) = $

If $y=y(x)$ satisfies the differential equation $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = (\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} dx$ for $x>0$ and $y(0)=\sqrt{7}$,then find $y(256)$.

Let a curve $y=y(x)$ be given by the solution of the differential equation $\cos \left(\frac{1}{2} \cos ^{-1}\left(e^{-x}\right)\right) d x=\sqrt{e^{2 x}-1} \,d y$. If it intersects the $y$-axis at $y=-1$,and the intersection point of the curve with the $x$-axis is $(\alpha, 0)$,then $e^{\alpha}$ is equal to $.....$

The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$ is

The solution of the equation $\frac{dy}{dx} = y(e^x + 1)$ is