# Let y=y(x) be the solution of the differential equation

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 $y\left(\frac{\pi}{4}\right)=0 .$ Then, the value of $(y(0)+1)^{2}$ is equal to :

1. $e^{1 / 2}$

2. $\mathrm{e}^{-1 / 2}$

3. $\mathrm{e}^{-1}$

4. e

Correct Option: , 3

Solution:

$\frac{d y}{d x}+2 \sin ^{2} x=1+y \cos 2 x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+(-\cos 2 \mathrm{x}) \mathrm{y}=\cos 2 \mathrm{x}$

I.F. $=e^{\int-\cos 2 x d x}=e^{-\frac{\sin 2 x}{2}}$

Solution of D.E.

$y\left(e^{-\frac{\sin 2 x}{2}}\right)=\int(\cos 2 x)\left(e^{-\frac{\sin 2 x}{2}}\right) d x+c$

$\Rightarrow \mathrm{y}\left(\mathrm{e}^{-\frac{\sin 2 \mathrm{x}}{2}}\right)=-\mathrm{e}^{\frac{\sin 2 \mathrm{x}}{2}}+\mathrm{c}$

Given

$y\left(\frac{\pi}{4}\right)=0$

$\Rightarrow 0=-\mathrm{e}^{-1 / 2}+\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{e}^{-1 / 2}$

$\Rightarrow \mathrm{y}\left(\mathrm{e}^{-\frac{\sin 2 \mathrm{x}}{2}}\right)=-\mathrm{e}^{\frac{\sin 2 \mathrm{x}}{2}}+\mathrm{e}^{-1 / 2}$

at $x=0$

$y=-1+e^{-1 / 2}$

$\Rightarrow \mathrm{y}(0)=-1+\mathrm{e}^{-1 / 2} \Rightarrow(\mathrm{y}(0)+1)^{2}=\mathrm{e}^{-1}$