Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 21

Given y'+y/x=0 and find the particular solution at y(2)=2
when the first order linear ordinary Differentian 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=0--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = 1/x and q(x)=0
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) *(0)) dx +c)/e^(int(1/x) dx)
first we shall solve
e^(int (1/x) dx)=e^(ln(x)) =x
so
proceeding further, we get
y(x) =((int e^(int (1/x) dx) *(0)) dx +c)/e^(int(1/x) dx)
=(0 +c)/(x)
=c/x
so y(x)=c/x
now let us find the value of c at y(2) =2
so,
y(x)=c/x
=> 2=c/2
=> c=4
so at y(2) =2 the particular solution is y(x) =4/x