The teacher hasn't uploaded a solution for this question yet.
The given differential equation is $\left[x\sin^{2}\left(\frac{y}{x}\right)-y\right]dx+x~dy=0$. We can rewrite this as $x\frac{dy}{dx} = y - x\sin^{2}\left(\frac{y}{x}\right)$ $\frac{dy}{dx} = \frac{y}{x} - \sin^{2}\left(\frac{y}{x}\right)$. Let $F(x,y) = \frac{y}{x} - \sin^{2}\left(\frac{y}{x}\right)$. Then $F(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \sin^{2}\left(\frac{\lambda y}{\lambda x}\right) = \frac{y}{x} - \sin^{2}\left(\frac{y}{x}\right) = F(x,y)$. Thus, the given differential equation is homogeneous.
Let $y = vx$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting this into the differential equation, we get $v + x\frac{dv}{dx} = v - \sin^{2}(v)$ $x\frac{dv}{dx} = -\sin^{2}(v)$ $\frac{dv}{\sin^{2}(v)} = -\frac{dx}{x}$
Integrating both sides, we have $\int \frac{dv}{\sin^{2}(v)} = \int -\frac{dx}{x}$ $\int \csc^{2}(v) dv = -\int \frac{dx}{x}$ $-\cot(v) = -\ln|x| + C$ $\cot(v) = \ln|x| - C$
Substituting $v = \frac{y}{x}$, we get $\cot\left(\frac{y}{x}\right) = \ln|x| - C$
Given that $y = \frac{\pi}{4}$ when $x = 1$. $\cot\left(\frac{\pi/4}{1}\right) = \ln|1| - C$ $\cot\left(\frac{\pi}{4}\right) = 0 - C$ $1 = -C$ $C = -1$
Substituting $C = -1$ into the general solution, we get $\cot\left(\frac{y}{x}\right) = \ln|x| - (-1)$ $\cot\left(\frac{y}{x}\right) = \ln|x| + 1$
Final Answer: $\cot\left(\frac{y}{x}\right) = \ln|x| + 1$
AI generated content. Review strictly for academic accuracy.