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?
Best Answer
As $y\equiv\pm 1$ are constant solutions, and the ODE is differentiable, the uniqueness theorem applies and no solution can cross any other, especially the constant ones.
On the intervals $(-\infty,-1),(-1,1), (1,\infty)$ the expression $\frac{y+1}{y-1}$ has a constant sign, so that this expression has constant sign on any solution of the ODE. This constant sign can be attached to the constant on the right side.