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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?

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    $\begingroup$ write $$e^{2x+C}=e^{2x}\cdot e^{C}$$ and set $$e^{C}=C'$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 29 '17 at 18:09
  • $\begingroup$ I don't see how that helps remove the absolute value sign. $\endgroup$ – Pame Nov 29 '17 at 18:25
  • $\begingroup$ you can write $$(-1)C'=C''$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 29 '17 at 18:26
  • $\begingroup$ 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. $\endgroup$ – Pame Nov 29 '17 at 18:53
  • $\begingroup$ 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. $\endgroup$ – Pame Nov 30 '17 at 9:43
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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.

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  • $\begingroup$ 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? $\endgroup$ – Pame Nov 30 '17 at 21:55
  • $\begingroup$ 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. $\endgroup$ – LutzL Nov 30 '17 at 22:08

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