# Integration (ln/e)

1. Sep 12, 2011

### Air

This is a question...

For the following question:
$y^{'}=\frac{dy}{dx}=3y$

I get the solution...
$\int \frac{1}{3y} dy = \int dx$
$\frac{1}{3}ln y = x + c$
$y = e^{3x}+e^{3c}$

However the textbook example says the solution is...
$y = ce^{3x}$

My question is would my answer be incorrect? How should the arbitrary constant be placed in ln and e integral solutions?

2. Sep 12, 2011

### micromass

Staff Emeritus
$e^{x+y}=e^xe^y$ and not $e^{x+y}=e^x+e^y$. You made this mistake in your last line.

3. Sep 12, 2011

### susskind_leon

You have to take into account that
$$e^{x+y}=e^x \cdot e^y \neq e^x + e^y$$
To get a feeling for that relation, take for example
$$2^{3+4}=(2 \cdot 2 \cdot 2 )\cdot (2 \cdot 2 \cdot 2 \cdot 2 )=2^3 \cdot 2^4$$

4. Sep 13, 2011

### Air

Oh yes, that is correct. Silly mistake. Thanks.