Question

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

• A. $e$
• B. $e^{1/2}$
• C. $e^{-1}$
• D. $e^{-1/2}$

Answer 1

The correct option is C $e^{-1}$

Given: ${\text{cosec}}^{2}xdy=({\text{cosec}}^{2}x-2)dx+\left(\cos 2x{\text{cosec}}^{2}x\right)ydx$
$\frac{dy}{dx}=(1-2{\sin }^{2}x)+\cos 2x.y$
$\frac{dy}{dx}-\cos 2x.y=\cos 2x$
$I.F$ is $e^{-\int \cos 2xdx}=e^{-\frac{\sin 2x}{2}}$

$\therefore y.e^{\frac{-\sin 2x}{2}}=\int \cos 2x.e^{\frac{-\sin 2x}{2}}dx$
$\Rightarrow y.e^{\frac{-\sin 2x}{2}}=-e^{\frac{-\sin 2x}{2}}+c\Rightarrow y=-1+c{e}^{\frac{\sin 2x}{2}}$
$y\left(\frac{\pi }{4}\right)=0\Rightarrow c=e^{\frac{-1}{2}}$
$\Rightarrow y=-1+e^{\frac{-1}{2}(1-\sin 2x)}$
$\Rightarrow y\left(0\right)=-1+e^{\frac{-1}{2}}$
$\Rightarrow \left(y\right(0)+1{)}^2=e^{-1}$