`xdx+ydy+[xdy-ydx]/[x^2+y^2]=0`

The solution of the differential equation xdx+ydy =(xdy-ydx)/(x^(2)+y^(2)) is tan(f(x, y)-C)=(y)/(x) (where, C is an arbitrary constant). If f(1, 1)=1 , then f(pi, pi) is equal to

xdy-ydx=xy^(2)dx

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

Solve xdx+ydy=xdy-ydx.

xdx + is + (xdy-ydx) / (x ^ (2) + y ^ (2)) = 0

The solution of (xdx+ydy)/(xdy-ydx)=sqrt((1-x^(2)-y^(2))/(x^(2)+y^(2))) is

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