Hello,

This question is very simple. Divide the equation by x. logx, on both sides.

So we get :

dy/dx + y/ x.logx  = 2. logx / x .logx

i.e. dy/dx + y/ x logx = 2/x

Clearly this is a linear differential equation. So firstly we find the integrating factor.

I.F. = e ^ ( intg. { 1/ xlogx dx } ) ,

I.F. = logx

Now the solution of this differential equation is given by

y. I.F = intg. ( Q. I.F dx ) where Q = 2/x

Hence solution is : y. logx = (logx)^2 + c

Hope it helps.