求微分方程通解

Question

(ysinx-1)dx-cosxdy=0
方法,步骤

Answer 1

Answer 2

(ysinx-1)dx-cosxdy=0

y'-ytanx=-secx
y=e^∫tanxdx{∫-secx[e^(-∫tanxdx)]dx+C}

Answer 3

ysinxdx-cosxdy -dx=0
-ydcosx-cosxdy=dx
(ydcosx+cosxdy)=-dx
d(ycosx)=-dx
∴ycosx=-x+C本回答被提问者采纳

Answer 4