The solution of the differential equation dy/dx + (y/x) = e^x is-

xy = e^x(x-1) +C

The given differential equation is dy/dx + (y/x) = e^x

which is of the form dy/dx + py =Q (Where p= 1/x and Q=e^x )

Now, I.F. = e^∫pdx

I.F. = e^∫(dx/x)

So. I.F. = e^logx = x.

The solution is given by: y.(I.F.) = ∫Q x(I.F.)dx +C

⇒ y.x = ∫(e^x x x) dx +C

so. xy = e^x(x-1) +C