【高数微分方程题目】

y'＝y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1＝y^2＋2(sinx-1)y＋(sinx-1)^2－cosx＝(y+sinx-1)^2－cosx
即y'＋cosx＝(y+sinx-1)^2
令u＝y＋sinx－1，则原微分方程化为du/dx＝u^2，通解是－1/u＝x＋C，回代yu＝y+sinx－1，得原微分方程的通解是y＝1－sinx－1/(x+C)

设t=y+sinx-1,则y=t+1-sinx,y'=t'-cosx,
原式化为:
t'-cosx=t^2-cosx
==>t'=t^2
==>dt/t^2=dx
==>t=1/(C-x)
==>y=1/(C-x)+1-sinx.

y'=y² + 2(sinx-1)y + (sinx)²-2sinx-cosx+1 = (y + sinx - 1)² - cosx
即
(y + sinx - 1)' = (y + sinx - 1)²
d(y + sinx - 1)/(y + sinx - 1)² = dx
积分得
-1/(y + sinx - 1) = x + C
y = -sinx + 1 - 1/(x+C)
这是通解