Exponentiating expression containing ln(abs(x))

Solution 1:

For the exercise in question, there will be no absolute value in the general solution because that is taken care of by the arbitrary constant of integration.

Given

$$\frac{dy}{dt}=4y(y+2),\,y(0)=6 $$

As you point out, this is separable.

Rewriting, we get

\begin{eqnarray} \left(\frac{1}{y}-\frac{1}{y+2}\right)\,dy&=&8\,dt\\ \ln\left\vert \frac{y}{y+2}\right\vert&=&8t+c\\ \left\vert\frac{y}{y+2}\right\vert&=&e^{8t+c}\\ \frac{y}{y+2}&=&\pm e^ce^{8t}\\ \frac{y}{y+2}&=&ke^{8t} \end{eqnarray}

Now there is no absolute value in the general solution since the constant of integration, $k$ can be either positive or negative.