The differential equation $\cot y \, dx = x \, dy$ has a solution of the form

If $y = y(x)$ is the solution of the differential equation $(1 + e^{2x}) \frac{dy}{dx} + 2(1 + y^2)e^x = 0$ and $y(0) = 0$,then $6(y'(0) + (y(\log_e \sqrt{3}))^2)$ is equal to

Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^{2}+y+1}{x^{2}+x+1} = 0$ is given by $(x+y+1) = A(1-x-y-2xy)$,where $A$ is a parameter.

The particular solution of the differential equation $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ when $x=e, y=e^2$ is

The solution of $\log \left(\frac{dy}{dx}\right) = 2x - 5y$ with the initial condition $y(0) = 0$ is:

The solution of $x^2 + y^2 \frac{dy}{dx} = 4$ is