The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y = \frac{x+3}{x+1}$,where $x > -1$,which passes through the point $(0,1)$. Then $y(1)$ is equal to:

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

If $\frac{dy}{dx} + \frac{3}{\cos^2 x} y = \frac{1}{\cos^2 x}$,$x \in \left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ and $y\left( \frac{\pi}{4} \right) = \frac{4}{3}$,then $y\left( -\frac{\pi}{4} \right)$ equals

The differential equation $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$ represents a family of curves given by:

If $x = x(y)$ is the solution of the differential equation $y \frac{dx}{dy} = 2x + y^{3}(y+1)e^{y}$ with the initial condition $x(1) = 0$,then $x(e)$ is equal to: