∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0
==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0 (等式兩端同除√(1+y²))
==>√(1+y²)dx+xydy/√(1+y²)-cosydy=0
==>√(1+y²)dx+xd(√(1+y²))-cosy)dy=0
==>d(x√(1+y²))=d(siny)
==>x√(1+y²)=siny+C (C是積分常數(shù))
∴原方程的通解是x√(1+y²)=siny+C (C是積分常數(shù)).