Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 11

To be able to evaluate the problem: sqrt(1-4x^2)y'=x , we express in a form of y'=f(x) .
To do this, we divide both sides by sqrt(1-4x^2) .
y'=x/sqrt(1-4x^2)
The general solution of a differential equation in a form of y'=f(x) can
be evaluated using direct integration. We can denote y' as (dy)/(dx) .
Then,
y'=x/sqrt(1-4x^2) becomes (dy)/(dx)=x/sqrt(1-4x^2)
This is the same as (dy)=x/sqrt(1-4x^2) dx
Apply direct integration on both sides:
For the left side, we have: int (dy)=y
For the right side, we apply u-substitution using u =1-4x^2 then du=-8x dx or (du)/(-8)=xdx .
int x/sqrt(1-4x^2) dx = int1/sqrt(u) *(du)/(-8)
Applying basic integration property: int c f(x) dx = c int f(x) dx .
int1/sqrt(u) *(du)/(-8) = -1/8int1/sqrt(u)du
Applying Law of Exponents: sqrt(x)= x^1/2 and 1/x^n = x^-n :
-1/8int1/sqrt(u)du=-1/8int1/u^(1/2)du
=-1/8int u^(-1/2)du
Applying the Power Rule for integration: int x^n= x^(n+1)/(n+1)+C .
-1/8int u^(-1/2)du =-1/8 u^(-1/2+1)/(-1/2+1)+C
=-1/8 u^(1/2)/(1/2)+C
=-1/8 u^(1/2)*(2/1)+C
= -2/8 u^(1/2)+C
= -1/4u^(1/2)+C or -1/4sqrt(u)+C
Plug-in u = 1-4x^2 in -1/4u^(1/2) , we get:
int1/sqrt(u) *(du)/(-8)=-1/4sqrt(1-4x^2)+C

Combining the results, we get the general solution for differential equation
( sqrt(1-4x^2)y'=x)
as:
y= -1/4sqrt(1-4x^2)+C