# Separable differential equations

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1. Jun 28, 2015

### hitemup

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I've highlighted two equations on the screenshot. How did it proceed from the first to the second? I'm actually confused with the absolute values. What is the idea behind getting rid of the first absolute value(1-5v^2) while keeping the second one(x)?

2. Jun 28, 2015

### RUber

So for starters, the manipulation is something like:
$-\frac 15 \ln | 1 - 5v^2 | = \ln |x| + c$
\begin{align*} \ln | 1 - 5v^2 | &= -5(\ln |x| +c)\\ &= -5( \ln |x| - \ln e^{c} )\\ &=-5 ( \ln \frac{|x|}{e^{c} })\\ &= \ln \left(\frac{|x|}{e^{c} }\right)^{-5} \\ &= \ln \left(\frac{e^{5c} }{|x|^5}\right) \end{align*}
Then, removing the absolute value from the left gives: $1-5v^2 = \pm \left(\frac{e^{5c} }{|x|^5} \right)$
So $C = \pm e^{5c}$
There only reason not to take out the absolute value from x that I can see is so that C is not dependent on x.

3. Jul 4, 2015

### Dr. Courtney

You get rid of the ln by exponentiating both sides.