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JEE MAIN_2021_
22-JULY-2021,S2_
Question
Let $y=y(x)$ be the solution of the differential equation $\operatorname{cosec}^2 x d y+2 d x=(1+y \cos 2 x) \operatorname{cosec}^2 x d x$, with $\mathrm{y}\left(\frac{\pi}{4}\right)=0$. Then, the value of $(\mathrm{y}(0)+1)^2$ is equal to :
Select the correct option:
A
$e^{1 / 2}$
B
$e^{-1 / 2}$
C
$e^{-1}$
D
e
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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