Question

Find the solution of $x{\sin }^2\frac{y}{x}dx=ydx-xdy$ which passes through the point $(1,\frac{\pi }{4})$.

Answer 1

Let $y=vx$. Then $\frac{dy}{dx}=v+\frac{xdv}{dx}$.

Therefore

$x{\sin }^2\left(\frac{y}{x}\right)dx=ydx-xdy$

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

$x{\sin }^{2}v=vx-x(v+\frac{xdv}{dx})$

${\sin }^{2}v=v-(v+\frac{xdv}{dx})$

${\sin }^{2}v=-\frac{xdv}{dx}$

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

$\frac{dx}{x}=-\text{cosec}{}^{2}vdv$

$\int \frac{dx}{x}=-\int \text{cosec}{}^{2}vdv$

$\ln x=\cot v+C$

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

Now $y=\frac{\pi }{4}$ for $x=1$

Hence

$\ln \left(1\right)=\cot \left(\frac{\pi }{4}\right)+C$

$0=C+1$

$C=-1$

Hence the particular solution is

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