We can use the integral of $\frac{1}{x}$ in order to solve a separable first-order linear equation like this:
$\frac{dy}{dt} + f(t) y = 0$
$
ln |y| = \left(-\int f(t)\,dt\right) + C
$
and then:
$y = \pm e^{\left(-\int f(t)\,dt\right) + C} = \pm e^{C} e^{-\int f(t)\,dt}$
As far as I know, we are able to remove the absolute value because $y=0$ it's a solution, and, thanks to Cauchy theorem, no solution of the ode may change its sign.
I'm okay with this, the problem is that I found some exercise on my textbook where the absolute value is removed, even if $y=0$ is not a constant solution of the ODE:
E.G.
$$
x' = \frac{3x-2}{t^2+1}
$$
$ \frac{1}{3} log|3x-2| = arctan(t) + c$
$ 3x – 2 = ce^{3arctan (t)}$ where $ c \neq 0$
Why it was possible to remove the absolute value here? Why $3x-2$ doesn't change its sign?
Best Answer
Indeed, if you want write it as $\ln(3x-2)$, you must consider the region $I: x\neq 2/3$ instead. Unless, the term $\ln(3x-2)$ doesn't make sense at all. I think, if the writer didn't noted the right region as $I$, or take a positive sign as you noted, probably he assumed the right region before or he wanted to take one 1-parameter of solutions. For example, in OE $$yy'=(y+1)^2$$ by solving we get $$\frac{1}{y+1}+\ln|y+1|=x+c$$ we just need to indicate $y\neq-1$.