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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?

Pame
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    write $$e^{2x+C}=e^{2x}\cdot e^{C}$$ and set $$e^{C}=C'$$ – Dr. Sonnhard Graubner Nov 29 '17 at 18:09
  • I don't see how that helps remove the absolute value sign. – Pame Nov 29 '17 at 18:25
  • you can write $$(-1)C'=C''$$ – Dr. Sonnhard Graubner Nov 29 '17 at 18:26
  • Yes, thats how you go about removing the absolute value sign like i described in my post. The problem is i don't see why it works. – Pame Nov 29 '17 at 18:53
  • I don't understand the absolute value sign well enough to figure that out. I know if you have something something like $|x| = 5$ Your answer would be x = $\pm 5& so the absolute value sign should matter even if whats on the right side is always positive. From what i understand about absolute value signs, they only matter as far as to what you're taking the absolute value of. For example if the absolute value sign was surrounding an always positive number you wouldn't need it. But in this case we are considering the sign in regard to whats on the right hand side. Thats what i find confusing. – Pame Nov 30 '17 at 09:43
  • Ok so if i understand things correctly: You can remove the absolute value bars when the expression on the right hand side is always negative, because you can't take the absolute value of something and get a negative number. So we put $K = \pm C$ because C can be either negative or positive, and we want our constant to be negative in this case. So when saying $K = \pm C$ we are basically saying $K = +C$ if C is negative, and $K = -C $ if C is positive? Am i on the right track? – Pame Nov 30 '17 at 13:30
  • Can you remove the absolute value bars if you allow for complex values? – johnnyb Feb 15 '21 at 14:51

2 Answers2

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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 $$ y(x)=\frac{\sinh(x)+y(0)\cosh(x)}{\cosh(x)+y(0)\sinh(x)}=\frac{y(0)+\tanh(x)}{y(0)\tanh(x)+1} $$

Lutz Lehmann
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  • How do you know what $\frac{y+1}{y-1}$ has a constant sign? Even though the expression on the right always has a constant sign equal to the sign of C (if i understand correctly), i don't see how we can deduce information about what the sign of the fraction will be. Wouldn't the absolute value sign matter as long as the expression on the right isn't always negative anyway? – Pame Nov 30 '17 at 21:55
  • For instance, if $y(0)\in(-1,1)$ then by uniqueness $y(t)\in(-1,1)$ for all $t\in\Bbb R$. This implies that the numerator is positive and the denominator negative, so that the whole fraction is negative for all times. || In your solution up to that point $C=e^c$ is always positive. – Lutz Lehmann Nov 30 '17 at 22:08
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It is equivalent to saying that if we choose the form

$$ y= C_1 \log \frac{1+x}{1-x}~$$

then $x$ should stay within the domain interval $ x\in (-1,1)$

Outside this interval (of two vertical asymptotes) the integral for $ y$ is complex, not real.

Narasimham
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