Differential equation integrating factor conceptual question.

[td21]

Homework Statement

Sorry asking similar quesion again about absolute value. You can read the attachment.
u(x) is the integrating factor. Why absolute value is omitted in the integration? and why the integrating factor is not "1/|x|", with the absolute sign

Homework Equations

The Attempt at a Solution

i thought we must use the abs sign and the solution is wrong?

 

[HallsofIvy]

The answer is exactly the same as to the question you asked in
https://www.physicsforums.com/showthread.php?t=474430

Since 1/x is not continuous at x= 0, it cannot be "continued" across x= 0. We must choose to restrict x to be positive or negative. If we choose to restrict x to be positive, we have 1/x. If we choose to restrict x to be negative, we have -1/x.

Since you use the integrating factor by multiplying the entire equation by it, you will get the same solution either way:

Taking the integrating factor to be 1/x, we get
$$\frac{1}{x}\frac{dy}{dx}- \frac{y}{x^2}= \frac{d}{dx}\left(\frac{1}{x}y\right)= 2+ \frac{1}{x}$$

Taking the integrating factor to be -1/x, we have
$$\frac{-1}{x}\frac{dy}{dx}+ \frac{y}{x^2}= -\frac{d}{dx}\left(\frac{1}{x}y\right)= -2- \frac{1}{x}$$
But multiplying both sides of that equation by -1 gives the same as the first equation.