Here, we may take f=xp−yq,g=z(xp+yq)−2xy so that
∂(x,p)∂(f,g)=2xy,∂(z,p)∂(f,g)=−x2p,∂(y,q)∂(f,g)=−2xy,∂(z,q)∂(f,g)=xyp
from which it follows that
[f,g]=xp(yq−xp)=0 (here [f,g] is the J, i.e. jacobian)
since xp=yp .
The equations are therefore compatible.
It is readily shown that p=y/z,q=x/z , so that we have to solve
which has solution
zdz=ydx+xdy
where c1 is a constant.
z2=c1+2xy
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