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

If $\frac{dy}{dx} = y + 3$,$y + 3 > 0$ and $y(0) = 2$,then $y(\log 2)$ is equal to

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x+x^2)(1-y+y^2)$, with the initial condition $y(0) = \frac{1}{2}$. Then $(2y(1) - 1)$ is equal to:

Let $S$ be the family of curves given by the general solution of the differential equation $\frac{y^2 e^{-1 / y}}{\sqrt{x}} dx - 2 \sec \sqrt{x} dy = 0$. Then the equation of the curve belonging to $S$ and passing through $(\pi^2, 1)$ is

If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \, dy = (1+2 \sin y) \, dx$ for $x > 0$, and $y(256)=\frac{\pi}{2}$, $y(49)=\alpha$, then $2 \sin \alpha$ is equal to:

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 $y(256)$.