Find the general solution of the differential equations:`xlogx(dx)/(dy)+y=2/xlogx`

Explanation

Given, `x log x(dy)/(dx)+y=(2)/(x)logx`
`implies (dy)/(dx)+(y)/(xlogx)=(2)/(x^(2))`
Comparing with differential equation `(dy)/(dx)+Py=Q`
`P=(1)/(xlogx)` and `Q=(2)/(x^(2))`
`I.F.=e^(intPdx)=e^(int(1dx)/(xlogx))`
Let `logx=timplies(1)/(x)=(dt)/(dx)implies(dx)/(x)=dt`
`:. I.F. =e^(int((1)/(t)dt)=e^(log|t|)=t=logx`
`:.` Solution of given differential equation,
`y*I.F.=intQ*I.F.dx+C`
`impliesy*logx=int(2)/(x^(2))logxx+C`
`implies ylogx=2int(logx*(1)/(x^(2)))dx+C`
`=2[logxint(1)/(x^(2))dx-int{(d)/(dx)(logx)int(1)/(x^(2))dx}dx]+C`
`=2[logx(-(1)/(x))-int{(1)/(x)*(-(1)/(x))}dx]+C`
`=2(-(logx)/(x)+int(1)/(x^(2))dx)+C`
`=2(-(logx)/(x)-(1)/(x))+C`
`implies ylogx=-(2)/(x)(1+kogx)+C`