Find the general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.

$A$ curve passing through $(2, 3)$ and satisfying the differential equation $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$ for $x > 0$ 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:

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1$,with the initial condition $y(0)=1$. Then,the value of $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ is equal to:

The general solution of $\left(\frac{dy}{dx}\right)^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}$ is

The particular solution of $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ when $x=2, y=1$ is