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

The differential equation is:

$\left[x \sin^2\left(\frac{y}{x}\right) - y\right]dx + x\,dy = 0$

Rearranging:

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

Since $d\left(\frac{y}{x}\right) = \frac{x\,dy - y\,dx}{x^2}$, we have $x\,dy - y\,dx = x^2\,d\left(\frac{y}{x}\right)$

$x \sin^2\left(\frac{y}{x}\right)\,dx + x^2\,d\left(\frac{y}{x}\right) = 0$

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Dividing by $x^2$:

$\frac{\sin^2\left(\frac{y}{x}\right)}{x}\,dx + d\left(\frac{y}{x}\right) = 0$

Let $v = \dfrac{y}{x}$:

$\frac{\sin^2(v)}{x}\,dx + dv = 0$

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Dividing by $\sin^2(v)$:

$\frac{dx}{x} + \csc^2(v)\,dv = 0$

Integrating:

$\displaystyle\int \frac{dx}{x} + \int \csc^2(v)\,dv = C$

$\ln|x| - \cot(v) = C$

$\ln|x| - \cot\left(\dfrac{y}{x}\right) = C$

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Applying $y = \dfrac{\pi}{4}$ when $x = 1$:

$\ln(1) - \cot\left(\dfrac{\pi/4}{1}\right) = C$

$0 - 1 = C$

$C = -1$

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The particular solution is:

$\ln x - \cot\left(\dfrac{y}{x}\right) = -1$

$\ln x = \cot\left(\dfrac{y}{x}\right) - 1$

$x = e^{\cot\left(\frac{y}{x}\right) - 1}$