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(C) Given the differential equation: $\operatorname{cosec}^{2} x \, dy + 2 \, dx = (1+y \cos 2x) \operatorname{cosec}^{2} x \, dx$.Dividing by $\opera

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$e^{-1}$

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(C) Given the differential equation: $\operatorname{cosec}^{2} x \, dy + 2 \, dx = (1+y \cos 2x) \operatorname{cosec}^{2} x \, dx$.Dividing by $\operatorname{cosec}^{2} x \, dx$,we get: $\frac{dy}{dx} + 2 \sin^{2} x = 1 + y \cos 2x$.Rearranging the terms: $\frac{dy}{dx} - y \cos 2x = 1 - 2 \sin^{2} x$.Since $1 - 2 \sin^{2} x = \cos 2x$,the equation becomes: $\frac{dy}{dx} - y \cos 2x = \cos 2x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\cos 2x$ and $Q(x) = \cos 2x$.The integrating factor ($I$.$F$.) is $e^{\int P(x) \, dx} = e^{\int -\cos 2x \, dx} = e^{-\frac{\sin 2x}{2}}$.The solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$.$y \cdot e^{-\frac{\sin 2x}{2}} = \int \cos 2x \cdot e^{-\frac{\sin 2x}{2}} \, dx + C$.Let $u = -\frac{\sin 2x}{2}$,then $du = -\cos 2x \, dx$.So,$y \cdot e^{-\frac{\sin 2x}{2}} = -\int e^u \, du + C = -e^{-\frac{\sin 2x}{2}} + C$.Given $y(\frac{\pi}{4}) = 0$,we have $0 = -e^{-\frac{\sin(\pi/2)}{2}} + C = -e^{-1/2} + C$,so $C = e^{-1/2}$.Thus,$y \cdot e^{-\frac{\sin 2x}{2}} = -e^{-\frac{\sin 2x}{2}} + e^{-1/2}$.At $x = 0$,$y \cdot e^0 = -e^0 + e^{-1/2}$,which gives $y(0) = -1 + e^{-1/2}$.Therefore,$(y(0) + 1)^{2} = (-1 + e^{-1/2} + 1)^{2} = (e^{-1/2})^{2} = e^{-1}$.

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

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