问题描述: (1+y^2)dx+(xy-根号下(1+y^2 ) cosy)dy=0 1个回答 分类:数学 2014-12-10 问题解答: 我来补答 ∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0 (等式两端同除√(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是积分常数)∴原方程的通解是x√(1+y²)=siny+C (C是积分常数). 展开全文阅读