Solve:
`ydx-xdy+logxdx=0`

Explanation

The given equation can be written as
`(dy)/(dx)-1/xy=1/xlogx`
`I.F.=e^(int-1/x)=e^(-logx) = 1/x`
Therefore, the solution is
`y/x=int1/x^(2)logxdx+c`……………..(1)
`therefore y/x=log_(e)x.(-1/x)-int(1/x)(-1/x)dx+C`
`therefore y/x=(-log_(e)x)/(x)-1/xdx+C`
Hence the required solution is `y+1+logx=cx`.