问题描述： ..微分方程... 1个回答 分类：数学 2014-11-05 问题解答： 我来补答 ∵y'=sin²(x-y+1) ==>dy/dx=sin²(x-y+1)==>1-dy/dx=1-sin²(x-y+1)==>(dx-dy)/dx=cos²(x-y+1)==>d(x-y)/dx=cos²(x-y+1)==>d(x-y+1)/dx=cos²(x-y+1)==>d(x-y+1)/cos²(x-y+1)=dx==>sec²(x-y+1)d(x-y+1)=dx==>tan(x-y+1)=x+C (C是积分常数)∴原方程的通解是tan(x-y+1)=x+C (C是积分常数). 展开全文阅读
