Question

solve ydx-xdy+logxdx=0

Answer 1

ydx - xdy = -logx dx

xdy - ydx = logx.dx

now divided by x² in both sides

( xdy -ydx)/x² = logx.1/x² dx
integrate both sides

we know, differentiation of
y/x =(xdy-ydx)/x²

so, integration of (xdy -ydx)/x² =y/x

second part logx.1/x² dx is integrated by integration by parts.
use this concept
then ,

logx.1/x² dx =- logx1/x -1/x

now,
y/x = -(logx +1)/x + C

where C is constant