(x+y)dx - xdy = 0 Solve the first-order differential equation by any appropriate method

Given (x+y)dx - xdy= 0
=> (x+y) - xdy/dx= 0
=> x+y-xy'=0
=> x+y=xy'
=> 1+y/x=y'
=> y' -y/x=1
when the first order linear ordinary differential equation has the form of
y'+p(x)y=q(x)
then the general solution is ,
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)
so,
y' -y/x=1--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = -1/x and q(x)=1
so on solving with the above general solution we get:
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx
=(int e^(int (-1/x) dx) *(1) dx +c)/e^(int (-1/x) dx)
first we shall solve
e^(int (-1/x) dx)=e^(-ln(x))=1/x     
so
proceeding further, we get
y(x) =(int e^(int (-1/x) dx) *(1) dx +c)/e^(int -1/x dx)
 =(int (1/x) *(1) dx +c)/(1/x)
=(ln(x) +c)/(1/x ) = x(ln(x)+c)
So , y(x) = x(ln(x)+c)