Let $$y = y(x)$$ be the solution of the differential equation $$\cosec^2 x \, dy + 2dx = (1 + y\cos 2x) \cosec^2 x \, dx$$, with $$y\left(\frac{\pi}{4}\right) = 0$$. Then, the value of $$(y(0) + 1)^2$$ is equal to:

Rewrite the differential equation $$\csc^2 x\,dy + 2\,dx = (1 + y\cos 2x)\csc^2 x\,dx$$ by moving all terms to one side: $$\csc^2 x\,dy = \bigl[(1 + y\cos 2x)\csc^2 x - 2\bigr]\,dx$$.

Dividing both sides by $$\csc^2 x\,dx$$ gives $$\frac{dy}{dx} = 1 + y\cos 2x - 2\sin^2 x$$. Using the identity $$2\sin^2 x = 1 - \cos 2x$$, this simplifies to $$\frac{dy}{dx} = 1 + y\cos 2x - 1 + \cos 2x = \cos 2x\,(1 + y)$$.

This is a separable equation. Separating variables: $$\frac{dy}{1 + y} = \cos 2x\,dx$$. Integrating both sides: $$\ln|1 + y| = \frac{\sin 2x}{2} + C$$.

Apply the initial condition $$y\!\left(\frac{\pi}{4}\right) = 0$$: $$\ln|1| = \frac{\sin(\pi/2)}{2} + C = \frac{1}{2} + C$$, so $$C = -\frac{1}{2}$$.

The solution is $$\ln|1 + y| = \frac{\sin 2x}{2} - \frac{1}{2}$$. At $$x = 0$$: $$\ln|1 + y(0)| = 0 - \frac{1}{2} = -\frac{1}{2}$$, giving $$1 + y(0) = e^{-1/2}$$.

Therefore $$(y(0) + 1)^2 = \left(e^{-1/2}\right)^2 = e^{-1}$$.