高等数学，求解微分方程通解？划圈那一步是怎么来的？求解答？

Question

如题所述

Answer 1

y'-ycotx =cscx
p(x) = -cotx
∫ p(x) dx
=∫ -cotx dx
=∫ -(cosx/sinx) dx
=-lnsinx +C
e^[∫ p(x) dx ] = 1/sinx
y'-ycotx =cscx
(1/sinx). [y'-ycotx] =(cscx).(1/sinx)
d/dx ( y/sinx) = (cscx)^2
∫ d(y/sinx)= ∫ (cscx)^2 dx
y/sinx = -cotx+C
y= (-cotx +C).sinx