Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 5

Given dy/dx +y/x=6x+2
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,
dy/dx +y/x=6x+2--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = 1/x and q(x)=6x+2
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) *(6x+2)) dx +c)/e^(int 1/x dx)
first we shall solve
e^(int (1/x) dx)=e^ln(x) =|x| as we know int (1/x)dx = ln(x)
so then we get e^(int (1/x) dx) =x
since x must be greater than 0, or elseln(x) is undefined.

Proceeding further with
y(x) =((int e^(int (1/x) dx) *(6x+2)) dx +c)/e^(int 1/x dx)
= (int x *(6x+2) dx +c)/x
= (int (6x^2 +2x) dx+c)/x
= (6x^3/3 +2x^2/2 +c)/x
= (2x^3+x^2+c)/x

So, y= (2x^3+x^2+c)/x