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

Comments

Popular posts from this blog

How does Bilbo show leadership and courage in The Hobbit?

What was the effect of World War II on African Americans?

How is Jack in William Golding's The Lord of the Flies presented as militaristic?