After performing indefinite integration on both sides, an equation becomes:

$$\ln(|v|) + C = \frac{-t}{RC} + C $$

This then simplifies to:

$$\ln(|v|) = \frac{-t}{RC} + C $$

Where did the $+ C$ go on the left hand side and why?

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    $\begingroup$ The first equation is technically incorrect because as written it implies that you have the same constant of integration from both integrals, which is not necessarily true. $\endgroup$ – David K Apr 2 at 3:52
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    $\begingroup$ We have three different numbers here that are all called the same name $C$. $\endgroup$ – littleO Apr 2 at 4:11
  • $\begingroup$ Right, sorry about the confusion. For verification, the C in the demoninator represents capacitance (for convenience you can think of this as any other arbitrary letter/number) and the + C on both sides is of course representative of the CoI. $\endgroup$ – threebeesdizzy Apr 2 at 4:34
  • $\begingroup$ I wish indefinite integrals and the +C were banned from calculus courses. They do more damage than good in learning the key concepts of calculus. Only definite integrals should ever be used. You’ll never see an indefinite integral or the +C again after you finish Calculus 2. $\endgroup$ – Deane Apr 2 at 4:56
  • $\begingroup$ And I do recommend changing a variable name if you find yourself using the same name for two different things. $\endgroup$ – Deane Apr 2 at 4:57

Let's say you have two functions $f(x)$ and $g(x)$, with $F(x)$ and $G(x)$ being their antiderivatives.

So, if you have $f(x) = g(x)$, and you integrate both sides, you get $F(x) + C_1 = G(x) + C_2$. Both $C$ are undetermined constants, because they ae the constants of integration, they can be any constant. Moving the constants over to one side, we get $F(x) = G(x) +C_2 - C_1$.

An undetermined constant minus an undetermined constant is just another undetermined constant, so we can replace it with one undetermined constant $C$. So, we get $F(x) = G(x) + C$.

  • $\begingroup$ Awesome. This makes perfect sense now. Thank you so much. $\endgroup$ – threebeesdizzy Apr 2 at 5:27
  • $\begingroup$ @threebeesdizzy You can hit the checkmark by the vote counter to accept this answer as the right one if it's the best one $\endgroup$ – Some Guy Apr 2 at 5:33

Two constants $C_1$ and $C_2$ can be represented as a single constant $C=C_1+C_2$ or $C=C_1-C_2$.

  • $\begingroup$ You're not explaining why there ought to be two constants, nor why you can represent them as one. I know the answer, but you should be explaining to the asker. $\endgroup$ – user21820 Apr 2 at 12:33

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