Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 18

The problem: 2xy'-ln(x^2)=0 is as first order ordinary differential equation that we can evaluate by applying variable separable differential equation:
N(y)y'=M(x)
N(y)(dy)/(dx)=M(x)
N(y) dy=M(x) dx
Apply direct integration: intN(y) dy= int M(x) dx to solve for the
general solution of a differential equation.
Then, 2xy'-ln(x^2)=0 will be rearrange in to 2xy'= ln(x^2)
Let y' = (dy)/(dx) , we get: 2x(dy)/(dx)= ln(x^2)
or2x(dy)= ln(x^2)(dx)
Divide both sides by x to express in a form of N(y) dy=M(x) dx :
(2xdy)/x= (ln(x^2)dx)/x
2dy= (ln(x^2)dx)/x
Applying direct integration, we will have:
int 2dy= int (ln(x^2)dx)/x
For the left side, recall int dy = y then int 2dy = 2y
For the right side, we let u =x^2 then du =2x dx or dx=(du)/(2x) .
int (ln(x^2))/xdx=int (ln(u))/x*(du)/(2x)
=int (ln(u)du)/(2x^2)
=int (ln(u)du)/(2u)
=1/2 int ln(u)/u du

Let v=ln(u) then dv = 1/udu ,we get:
1/2 int ln(u)/u du=1/2 int v* dv
Applying the Power Rule of integration: int x^n dx = x^(n+1)/(n+1)+C
1/2 int v* dv= 1/2 v^(1+1)/(1+1)+C
= 1/2*v^2/2+C
=1/4v^2+C
Recall v = ln(u) and u = x^2 then v =ln(x^2) .
The integral will be:
int (ln(x^2))/xdx=1/4(ln(x^2))^2 +C or(ln(x^2))^2 /4+C
Combing the results from both sides, we get the general solution of the differential equation as:
2y = (ln(x^2))^2 /4+C
or y =(ln(x^2))^2 /8+C

To solve for the arbitary constant (C), we consider the initial condition y(1)=2
When we plug-in the values, we get:
2 =(ln(1^2))^2 /8+C
2 =0/8+C
2=0+C
then C=2
.Plug-in C=2 on the general solution: y =(ln(x^2))^2 /8+C , we get the
particular solution as:
y =(ln(x^2))^2 /8+2