Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 23

Given xdy = (x + y + 2)dx
=> xdy/dx = (x + y + 2)
=> y'=1+y/x+2/x
=> y'-y/x = 1+2/x
=> y'-y/x =(x+2)/x
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 =(x+2)/x--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = -1/x and q(x)=(x+2)/x
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) *((x+2)/x)) dx +c)/e^(int (-1/x) dx)
first we shall solve
e^(int (-1/x) dx)=e^(ln(1/x)) = (1/x)
so
proceeding further, we get
y(x) =((int e^(int (-1/x) dx) *((x+2)/x)) dx +c)/e^(int (-1/x) dx)
=((int (1/x) *((x+2)/x)) dx +c)/(1/x)
=(int (1/x) *((1+(2)/x)) dx +c)/(1/x)
= x(ln(x)-2/x +c)
so now let us find the particular solution of differential equation at y(1)=10
y(1) = 1(ln(1)-2/1 +c)
=> 10 = 0-2+c
c=12
y(x) =x(ln(x)-2/x +12) is the solution