`y^2(xdy+ydx)+xdy-ydx=0`

The curve,which satisfies the differential equation (xdy-ydx)/(xdy+ydx)=y^(2)sin(xy) and passes through (0,1), is given by

xdy-ydx = yes

xdy-ydx=xy^(2)dx

Solve x^2dy+y^2dx=xy(xdy-ydx)

y log ydx-xdy=0

dx-dy+ydx+xdy=0

x^(2)y(2xdy+3ydx)=dy

Solve: xdy+ydx= (xdy-ydx)/(x^(2)+y^(2))

Solve: xdy+ydx= (xdy-ydx)/(x^(2)+y^(2))