Question

find the solution of the following differential equation ` x logx (dy)/(dx) + y = 2 log x`

Answer 1

`dy/dx+y/(xlogx)=2/x`
`P(x)=1/(xlogx)`
`Q(x)=2/x`
`IF=e^(loglogx)=logx`
`y*IF=intlogx*2/x dx`
`y*logx=int tdt`
`ylogx=t^2/2*2+c`
`ylogx=(logx)^2+c`
`y=logx+2/logx`.