xdy/dx=ylny，

dy/(ylny)=dx/x，

ln(lny)=lnx+lnC，

lny=Cx，

Then the general solution of the differential equation xy'-ylny=0 is y = e (CX), where c is an integer constant.

(2)sinxdy+cosydx=0

= = & gtdy/cosy+dx/sinx=0

= = & gtdy/cosy+∫dx/sinx=0

= = & gtln │ secy+tany │-ln │ cscx+cotx │ = ln │ c │ (c is a non-zero constant)

= = & gt(secy+ Tani) /(cscx+cotx)=C

= = & gtsecy+ Tani =C(cscx+cotx)

The general solution of this equation is secy+tany=C(cscx+cotx).

(3) xy'=y(lny-lnx)

xy'/y=lny-lnx

x(lny)'=lny-lnx

x(lny)'-lny=-lnx

[x(lny)'-lny]/x？ =-(lnx)/x？

[(lny)/x]'=-(lnx)/x？

Two-sided integrated:

(lny)/x= - ∫(lnx)/x？ Advanced (short for deluxe)

=∫ (lnx) d( 1/x)

=(lnx)/x-∫ ( 1/x) d(lnx)

=(lnx)/x-∫( 1/x)*( 1/x)dx

=(lnx)/x+ 1/x+C

So:

lny=lnx+Cx+ 1