(dy/dx)sin x=yln y的通解

分离变量，dy/(yln y)=dx/sin x,∫dy/(yln y)= ∫ dx/sin x ，
∫dlny/ln y= ∫ cscxdx,ln(lny)=ln|cscx-cotx|+C1=ln|tan(x/2)|+C1,
(lny)=e^[ln|tan(x/2)|+C1],y=e^[e^(C1)|tan(x/2)|]=e^[Ctan(x/2)]

(dy/dx)sinx=ylny d(lny)/lny=sinxdx/[1-(cosx)^2]
ln[ln(y)]=-dcosx/[(1-cosx)(1+cosx)] ln[ln(y)]=(-1/2)[dcosx/(1-cosx)+dcosx/(1+cosx)]
ln[ln(y)=-ln√[(1-cosx)/(1+cosx)]+C
ln[(lny)]=lntg(x/2)+C

∫1/y*1/lny dy=∫1/sinx dx
lnlny=∫1/2/[sin(x/2)*cos(x/2)] dx
lnlny=ln(sin(x/2))-ln(cos(x/2))+c
lny=e^c*tan(x/2) 这里e^c写作C,因为毕竟还是常数.
y=e^(C*tan(x/2))