# Thread: Dropping the Absolute Value

1. ## Dropping the Absolute Value

I was helping someone today on a basic Differential Equations problem and the answer I got disagreed with the one given in the back of the book. I'm hoping someone can confirm which is correct:

The problem is:
dy/dx = y/x

The answer I got (after combining all constants into K) is:
y = K|x|

The answer in the back of the book is:
y = Kx

(NOTE: There were no initial limits put on the domain or range of the function.)

I'm pretty sure the book "cheated" and decided to drop the absolute value (assuming the + and - would be absorbed into the K).
Can someone confirm whether that's legal to do?

2. I got ln y = ln |kx|, then y = kx (equalin' logarithm quantities)

Remember that "k" is an arbitrary constant.

That's what I think.

3. Originally Posted by ecMathGeek
I was helping someone today on a basic Differential Equations problem and the answer I got disagreed with the one given in the back of the book. I'm hoping someone can confirm which is correct:

The problem is:
dy/dx = y/x

The answer I got (after combining all constants into K) is:
y = K|x|

The answer in the back of the book is:
y = Kx

(NOTE: There were no initial limits put on the domain or range of the function.)

I'm pretty sure the book "cheated" and decided to drop the absolute value (assuming the + and - would be absorbed into the K).
Can someone confirm whether that's legal to do?
Originally Posted by Krizalid
I got ln y = ln |kx|, then y = kx (equalin' logarithm quantities)

Remember that "k" is an arbitrary constant.

That's what I think.
What counts in the final analysis is what the original differential equation says. I see nothing to restrict x and y from being negative there, so I see no reason to restrict the solution.

I'm not too sure of the theory, but I would suggest that we might be able to "get rid" of the absolute value bars by solving the equation TWICE: once on (-infinity, 0] and [0, infinty) making sure that the two solutions match at x = 0. Someone better versed in theory than me would need to verify this, though.

-Dan

4. Originally Posted by ecMathGeek
I was helping someone today on a basic Differential Equations problem and the answer I got disagreed with the one given in the back of the book. I'm hoping someone can confirm which is correct:

The problem is:
dy/dx = y/x
We are solving this on some open interval I \ {0}.
That is,
(0,N) where N is finite or infinite.
Or,
(-N,0) where N is finite or infinite.

In that case, these are the the solutions:
y=0 on I
Or,
ln |y| = ln|x|+C
Thus,
exp (ln|y|) = exp (\ln|x|+C)
|y| = |x|*exp(C)=K*|x| where K>0

Case 1: I=(0,N)
|y|=K*|x|=Kx
Thus,
y=+/- K*x = Mx where M is non-zero real.
Together with trivial solution y=0.
We have,
y=Mx where M is any real number.

Case 2: I=(-N,0)
|y|=K*|x|=-Kx
y=+/- (-K)x=Mx where M is non-zero real.
Together with trivial solution y=0.
We have,
y=Mx where M is any real number.

In both cases both solutions work.
Thus ye book did not cheat.
(However, it was not nice because they did not specify the interval on which to find the solution).

5. Originally Posted by Krizalid
I got ln y = ln |kx|, then y = kx (equalin' logarithm quantities)

Remember that "k" is an arbitrary constant.

That's what I think.
You cannot really do not (though it produces the right answer)
Becuase,

ln x + C= ln x + ln k = ln (xk) only works if C>0

6. Originally Posted by ecMathGeek

From this, I get 4 answers:
y = Kx ... {(x,y) | x>=0, y>=0}
y = -Kx ... {(x,y) | x<=0, y>=0}
-y = Kx ... {(x,y) | x>=0, y<=0}
-y = -Kx ... {(x,y) | x<=0, y<=0}
You only work with "x" not with "y", because you can determine "y" once you know "x".

[Edit: I wrote all this before I saw what THP had written. Thanks again ThePerfectHacker.]
My foot. All of the sudden when I type the answer you realize the answer.

7. Sorry, my Moderator Edit button is located next to the Quote Button. I deleted your entire post.

I wanted to add, that do not worry how differencial equations work. Typical Textbooks do not do much theory. But it is complicated, very complicated. Thus, they just avoid it. You will no way be asked on your exam.

8. Originally Posted by ecMathGeek
You only work with "x" not with "y", because you can determine "y" once you know "x".

My foot. All of the sudden when I type the answer you realize the answer.
It's funny that I would be quoting myself. But, yes, in fact I did realize the answer as you were typing it. This may surprise you, but I do actually know math pretty well. Why would I lie about having figured out the answer?

9. Originally Posted by ecMathGeek
This may surprise you, but I do actually know math pretty well.
doesn't surpirse me!

10. Originally Posted by ThePerfectHacker
Sorry, my Moderator Edit button is located next to the Quote Button. I deleted your entire post.

I wanted to add, that do not worry how differencial equations work. Typical Textbooks do not do much theory. But it is complicated, very complicated. Thus, they just avoid it. You will no way be asked on your exam.
I appreciate you trying to look out for me, but it's not necessary. I'm a math major - if I'm content not knowing how the math works then what the heck am I doing majoring in math? I did not create this post because I am having a hard time understanding Differential Equations (especially equations as simple as the one I posted) but because I wanted to know if an absolute value could be removed from the final answer. I'm not having a hard time at all understanding Differential Equations.

I realize that many of the theories behind Differential Equations are given by proofs that are beyond my current level of understanding, but I am not content not knowing simply because I am not yet able to understand.

11. Originally Posted by ecMathGeek
I realize that many of the theories behind Differential Equations are given by proofs that are beyond my current level of understanding, but I am not content not knowing simply because I am not yet able to understand.
I have the same problem you have. But I agree with the approach colleges teach it.

My philosophy in mathematics is first learn how do it (countour integrations, differencial equations, line integrals...). Then when you get good at the stuff then learn the theory behind it. That is exactly what "Advanced Claculus" is. First you spend close to 2 years learning calculus and then finally when you know it you can start learning why it works.

This approach to math is not always true. If you are learning some area in mathematics that is completely based on none-stop proofs (like set theory) then no you cannot say the above.