Solve `x(dy)/(dx)=y(logy-logx+1)`

Explanation

Given, `" "x(dy)/(dx)=y(logy-logx+1)`
`rArr" "x(dy)/(dx)=ylog((y)/(x)+1)`
`rArr" "(dy)/(dx)=(y)/(x)(log""(y)/(x)+1)" "`...(i)
which is a homogeneous equation.
Put `" "(y)/(x)=v or y=vx`
`therefore" "(dy)/(dx)=v+x(dv)/(dx)`
On substituting these values in Eq. (i), we get
`rArr" "x(dv)/(dx)=v(logv+1-1)`
`rArr" "x(dv)/(dx)=v(logv)`
`rArr" "(dv)/(vlogv)=(dx)/(x)`
On integrating both sides, we get
`" "int(dv)/(vlogv)=int(dx)/(x)`
On putting logv=u in LHS integral, we get `" "(1)/(v)*dv=du`
`" "int(du)/(u)=int(dx)/(x)`
`rArr" "logu=logx+logC`
`rArr" "logu=logCx`
`rArr" "u=Cx`
`rArr" "logv=Cx`
`rArr" "log((y)/(x))=Cx`