`y dx+x log((y)/(x))dy-2x dy=0`

Explanation

`y dx+x log((y)/(x))dy-2x dy=0`
`implies y dx+[x log((y)/(x))-2x]dy=0`
`implies (y)/(x)+[log((y)/(x))-2](dy)/(dx)=0`
`implies (dy)/(dx)=[((y)/(x))/(2-log((y)/(x)))]" ....(1)"`
माना y = vx (`because` समीकरण समघातीय है)
`implies (dy)/(dx)=v+x(dv)/(dx)`
समीकरण (1) में रखने पर,
`v+x(dv)/(dx)=(v)/(2-logv)`
`implies x(dv)/(dx)=(v)/(2-logv)-v`
`=(vlogv-v)/(2-logv)`
`implies (2-logv)/(v(logv-1))dv=(dx)/(x)`
`impliesint(2-logv)/(v(logv-1))dv=int(dx)/(x)`
`log v = t implies (1)/(v)dv=dt`
`implies int(2-t)/(t-1)dt=int(dx)/(x)`
`implies int(-1+(1)/(t-1))dt=int(dx)/(x)`
`implies -t+log(t-1)=logx+logc`
`implies-logv+log(logv-1)=log(cx)`
`implies(logv-1)/(v)=cx`
`implies logv-1=cxv`
`implies log((y)/(x))-1=cy`