The particular solution of the differential equation $\left\{x \sin^2 (\frac{y}{x}) - y\right\} dx+xdy = 0, y =\frac{\pi}{4}$ when $x = 1$ is

$x = e^{\cot\frac{y}{x} -1}$

The correct answer is Option (3) → $x = e^{\cot\frac{y}{x} -1}$

$\{x\sin^{2}(y/x)-y\}\,dx+x\,dy=0$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\sin^{2}\!\left(\frac{y}{x}\right)$

Let $v=\frac{y}{x}\Rightarrow y=vx,\ \frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=v-\sin^{2}v\ \Rightarrow\ x\frac{dv}{dx}=-\sin^{2}v$

$\displaystyle \int \csc^{2}v\,dv=-\int \frac{dx}{x}\ \Rightarrow\ -\cot v=-\ln x+C$

$\Rightarrow \cot v=\ln x+C\ \Rightarrow\ \cot\!\left(\frac{y}{x}\right)=\ln x+C$

Condition: $x=1,\ y=\frac{\pi}{4}\Rightarrow \cot\!\left(\frac{\pi}{4}\right)=1=\ln 1+C\Rightarrow C=1$

$\cot\!\left(\frac{y}{x}\right)=1+\ln x$