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I'm trying to understand a exercise about differential equations
$x'=\frac{1}{2}x+1$
I'm going for the general solution by using separable equation. Everything goes well until I get off the rails:
$\int\frac{1}{\frac{1}{2}x+1}=2*log(\frac{x+2}{2})+C$
I checked the step with Wolfram Alpha step by step solution and they're doing a step a absolutly can't understand and they come up with this:
$y_{(x)}= 2*log(x+2)+C$
What happend there? Why does the fraction inside the logarithm disappear? What rule was there applied?

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  • $\begingroup$ Ignoring the scalar, $\log((x+2)/2) = log(x+2) - \log(2)$, $-\log(2)$ is a constant, so roll it up into the $C$. $\endgroup$ – Echan Jul 13 '15 at 20:58
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Both solutions are correct, because:

$$2\log(\frac{x+1}{2})=2\log(x+1)-2\log 2$$

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  • $\begingroup$ Okay Thanks. This + Echan's answer has opened my eyes. $\endgroup$ – mathstud Jul 13 '15 at 21:04

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