..微分方程...

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 15:53:49

..微分方程...
..微分方程...

..微分方程...
∵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).

令u=x-y+1则du/dx=1-y'=1-(sinu)^2=(1+cos2u)/2
积分得u=(1/2)x + (1/4)sin2x + C
即x-y+1=(1/2)x + (1/4)sin2x+C
所以y=(1/2)x - (1/4)sin2x + C