When using the method of separating variables when solving differential equations you often end up with an absolute value around y. For example:

$\frac {dy}{dt} = ky $

$\frac {dy}{y} = k dt$

After integrating both sides you get:

$ ln(|y|) = kt + C$

$|y| = Ce^{kt}$

Now I want to remove the absolute value around Y, how do I do that without breaking any mathematical rules?


Note: My previous answer was wrong. $C$ mustn't always be positive. Though the differential equation derivation leads to a positive RHS, it's because of the absolute value bars that the RHS is positive.

The absolute value here can abstract away negative solutions. That means, you'll have to split it into two possibilities:

$$y = Ce^{kt}$$ $$y = -Ce^{kt}$$

Thus, to remove the absolute value bars, rewrite as so:

$$y = \pm Ce^{kt}$$

  • $\begingroup$ It's ok, just edit your answer at the end. No need for points desperately :-) cheers $\endgroup$ – Tal-Botvinnik Mar 11 '18 at 17:40
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    $\begingroup$ @Tal-Botvinnik Done, thanks for you comments! $\endgroup$ – Andrew Li Mar 11 '18 at 17:50

Hint: the expression of the right has the same sign of $C$, which is constant

  • $\begingroup$ Yeah, that's basically what i've read but I don't really understand it. Can I just lose the absolute value around Y with the motivation that the other side depends on C? $\endgroup$ – gbgult Mar 11 '18 at 15:02
  • $\begingroup$ @AndrewLi yeah, I get that too, but that doesn't really explain why I can drop the absolute value for me. If C is negative, would't we want the absolute value then? $\endgroup$ – gbgult Mar 11 '18 at 15:07
  • $\begingroup$ Be wary of posting "answers" that are significantly shorter than the Question to which they respond. I don't see such an immediate application of the hint that additional explanation can or should be omitted. Spell out your idea and show that it leads to a solution. $\endgroup$ – hardmath Mar 11 '18 at 15:50
  • $\begingroup$ Well, $C=y(0)$ so it depends on how it is defined... Note that the other answer claims that $C$ is always positive, which is not true! For instance $y=-e^{kt}$ is a perfectly ok solution to your equation. Beware of accepting answers just because they are long. My hint was ok, I don't see why the downvote. $\endgroup$ – Tal-Botvinnik Mar 11 '18 at 17:05

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