When solving the differential equation $$ y' = 1-y^2 $$ you get the solution $$ |\frac{y+1}{y-1}| = Ce^{2x} $$ You can then remove the absolute value sign by changing C to a new konstant $K = \pm C$. But why is this? I've been struggling really hard to grasp this concept, and I'm also finding it hard to have an intuitive understanding of what the absolute value sign actually means practically in this context. What would would be the difference between having the absolute value sign surrounding our fraction and it not being there?
Also, I've been told the the same differential equation also has the two constant solution $K = \pm 1$. From what i understand constant solutions are found by setting $Y = K$, but what do they actually mean, and what do you do if there is an x in the equation?
The equation is autonomous and scalar, $y'=f(y)$. Any roots $y_*$ of $f$ give constant solutions of the differential equations, as then both sides of it are constant zero.
As $y(x)\equiv\pm 1$ are constant solutions, and the ODE is differentiable, the uniqueness theorem applies and no solution can cross any other, especially the constant ones. Considering the two constant solutions, the the state space or line is split into 3 regions that are invariant, any solution starting in such a region stays in that region.
On the intervals $(-\infty,-1),(-1,1), (1,\infty)$, if a solution starts at $y(0)$ in one of these intervals, it stays in this same interval in both time directions. The expressions $y+1$, $y-1$ and thus also in combination $\frac{y+1}{y-1}$ will have a constant sign, depending on the interval.
Or in other words, in the equation $$ \frac{y(x)+1}{y(x)-1}e^{-2x}=\pm C $$ on the left side is a continuous function. This means that the right side can not jump in its sign, it is just a constant. Thus also the left side is constant. The value and thus the sign of this constant is fixed by the initial conditions.
Completely alternatively, one can treat this as Riccati equation, set $y=\frac{u'}{u}$ to find $u''=u$, $u(0)=1$, $u'(0)=y(0)$, so that after inserting back $$ y(x)=\frac{\sinh(x)+y(0)\cosh(x)}{\cosh(x)+y(0)\sinh(x)}=\frac{y(0)+\tanh(x)}{y(0)\tanh(x)+1} $$