Solve $2(y + 3) - xy \frac{dy}{dx} = 0$, given that $y(1) = -2$.

$y \sin x = -\frac{1}{2} \cos 2x + \frac{3}{2}$

The correct answer is Option (4) → $y \sin x = -\frac{1}{2} \cos 2x + \frac{3}{2}$ ##

Given differential equation,

$dy = \cos x (2 - y \csc x) dx$

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

$\Rightarrow  \frac{dy}{dx} = 2 \cos x - y \csc x \cdot \cos x$

$\Rightarrow  \frac{dy}{dx} = 2 \cos x - y \cot x$

$\Rightarrow  \frac{dy}{dx} + y \cot x = 2 \cos x$

Which is a linear differential equation.

On comparing it with $\frac{dy}{dx} + Py = Q$, we get

$P = \cot x, \quad Q = 2 \cos x$

$\text{I.F.} = e^{\int P dx} = e^{\int \cot x dx} = e^{\log \sin x} = \sin x \quad [∵e^{\log x} = x]$

The general solution is

$y \cdot \sin x = \int 2 \cos x \cdot \sin x , dx + C \quad [y \times \text{I.F.} = \int (Q \cdot \text{I.F.}) dx + C]$

$\Rightarrow \quad y \cdot \sin x = \int \sin 2x , dx + C \quad [∵\sin 2x = 2 \sin x \cos x]$

$\Rightarrow \quad y \cdot \sin x = -\frac{\cos 2x}{2} + C \quad \dots(i)$

When $x = \frac{\pi}{2}$ and $y = 2$, then

On putting the value of $x$ and $y$ in Eq. (i), we get

$2 \cdot \sin \frac{\pi}{2} = -\frac{\cos \left( 2 \times \frac{\pi}{2} \right)}{2} + C $

$\Rightarrow \quad 2 \cdot 1 = +\frac{1}{2} + C $

$\Rightarrow \quad 2 - \frac{1}{2} = C \Rightarrow \frac{4 - 1}{2} = C $

$\Rightarrow \quad C = \frac{3}{2}$

On substituting the value of $C$ in Eq. (i), we get

$y \sin x = -\frac{1}{2} \cos 2x + \frac{3}{2}$