$y+x^2=\frac{dy}{dx}$ has the solution

The integrating factor of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is . . . . . . .

If $y(x)$ is the solution of the differential equation $\frac{dy}{dx} + \left( \frac{2x + 1}{x} \right)y = e^{-2x}, x > 0$ where $y(1) = \frac{1}{2}e^{-2}$,then:

If $y=y(x)$ is the solution of the equation $e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x$ with $y(0)=0$,then $1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)$ is equal to

If a curve passes through the point $(1, -2)$ and has the slope of the tangent at any point $(x, y)$ on it as $\frac{x^2 - 2y}{x}$,then the curve also passes through the point:

Let $x=x(y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$ such that $x(1)=0$. Then,$x(e)$ is equal to