The general solution of $\frac{dy}{dx} = 2xye^{x^2}$ is

Let $y=y(x)$ be the solution of the differential equation $dy=e^{\alpha x+y} dx$; $\alpha \in N$. If $y(\log_{e} 2)=\log_{e} 2$ and $y(0)=\log_{e}(\frac{1}{2})$,then the value of $\alpha$ is equal to $.....$

The particular solution of the differential equation $x \, dy + 2y \, dx = 0$, given that $y = 1$ when $x = 2$, is:

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

Let $x = x(y)$ be the solution of the differential equation $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$,$y > 0$ and $x(1) = \frac{\pi}{2}$. Then $\cos(x(2))$ is equal to:

If $(2+\sin x) \frac{dy}{dx}+(y+1) \cos x=0$ and $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to