Understanding Absolute Values.

[jtabije]

Hey Guys! I've frequently come by this forum and have finally joined it in hopes that I could get some more conceptual insight in understanding math.

One thing that I have trouble with is absolute values. I understand that:

|x|= $$\sqrt{x^2}$$ .. and how it can be defined given restrictions on x.

..but I'm having some trouble trying to completely understand and confidently use them in some contexts.

For example, consider this simple first-order linear differential equation:

xy' + y = $$\sqrt{x}$$

Assuming you don't do this through inspection, you would get an integrating factor I such that:

I = e^ln(|x|) = |x|

How would one utilize that to find the solution given no bounds and restrictions on x?

 

[mathman]

One way would be to consider x > 0 (x = |x|) and x < 0 (x = -|x|) as separate cases.

 

[jtabije]

Thanks for the reply, Mathman. Is there another way than to separate it into separate cases?

I feel as if I'm forgetting a key concept. For example, consider this simplifcation that was given in a solution manual of mine:

I(x) = e^$${\int(-1/x dx)}$$ = e$$^{(-ln(x))}$$= e$$^{ln(x^{-1})}$$ = 1/x

Shouldn't it simplify to 1/|x|?

 

[Mark44]

Yes, since
$$\int \frac{dx}{x} = \ln{|x| + C$$

(I have omitted the negative sign in your problem to focus on the integral.)

If the context of this problem is a differential equation with an initial condition, the sign of the initial value is often used to choose positive values for x, or negative values.

 

[jtabije]

Ahh! The initial value! How could I have overlooked that, Mark44?

Thanks!

 

[HallsofIvy]

We might also point out that, in this particular case,
$$xy'+ y= \sqrt{x}$$
with y a real valued function, x cannot be negative so we would use |x|= x.