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

Answer

`y=xe^(cx), c gt 0`

Explanation

`x(dy//dx)=y(logy-logx+1)`
or `(dy)/(dx)=y/x[logy/x+1]`
Putting `y=vx`, we get `(dy)/(dx)=v+x(dv)/(dx)`
and the given equation transforms to
`v+x(dv)/(dx)=v[logv+1]`
or `x(dv)/(dx)=vlogv`
or `int(dv)/(dx)=vlogv`
or `int(dv)/(vlogv)=int(dx)/x`
or `loglogv=logx+logc, c gt 0`
or `cx=log(y//x)`
or `y=xe^(cx), c gt 0`