where does the modulus go when cancelling $e$ and $\ln$ in this problem?

Question

So I did this problem today:

Show that $\frac{dy}{dx} = yx^2$ can be written as $y = Ae^{\frac{x^3}{3}}$

my solution is shown below:

$$ \frac{dy}{dx} = yx^2 $$ $$ \frac{1}{y} dy = x^2 dx $$ $$ \int\frac{1}{y}\ dy = \int x^2\ dx $$ $$ \ln |y| = \frac{x^3}{3} + C $$ $$ e^{\ln |y|} = e^{\frac{x^3}{3} + C} $$ $$ y = e^{\frac{x^3}{3} + C} $$ $$ y = Ae^{\frac{x^3}{3}} $$

But I don't understand what happens to the modulus around the $y$? Why isn't it $|y| = Ae^{\frac{x^3}{3}}$? What happens if the modulus is in fact left there?

Answer 1

Because $e^x\ge 0\,\forall x\in\mathbb{R}$ so the modulus sign is redundant.

Answer 2

you have to remember, that $e^c$ respectively A is always positiv. So the whole Expression on the right hand is positiv: |y|=y

greetings,

calculus .