Differential equation, modulus signs in solution?

Question

Question: Find the equation of the curve with gradient $frac{dy}{dx}=frac{y+1}{x^2-1}$ that passes through $(-3,1)$.
So I integrated both sides with respect to $x$ which gave me
$ln{|y+1|}=frac{1}{2}ln{|x-1|}-frac{1}{2}ln{|x+1|}+c$
Given the point I found $c=ln{sqrt{2}}$
So at this point I figure the best way this equation can tidy up is $|y+1|=sqrt{frac{2|x-1|}{|x+1|}}$
I'm having trouble getting my head around what I can do with the equation at this point. The textbook gives an answer of $(y+1)^2(x+1)=2(x-1)$ but where have the modulus signs gone? Surely changing them to brackets gives just one possible curve (ie. where $y>-1$ and $x>1$)? And can you just square both sides of the equation like this? ($y=x$ and $y^2=x^2$ aren't the same curve)
Can anyone help me out with perhaps a little intuition here, and maybe some advice with how the solution should be presented here? Thanks in advance.

Lutz Lehmann · Answer

Any sign factor is a discrete property of a continuous function, thus constant, thus determined by the initial value. Thus to have
$$
pm(y+1)=sqrt{pm2,frac{x-1}{x+1}}
$$
defined and valid at the initial point one needs the positive sign in both places, so that
$$
y(x)=-1+sqrt{2,frac{x-1}{x+1}}.
$$

JJacquelin · Answer

$$frac{dy}{dx}=frac{y+1}{x^2-1}$$

$$ln{|y+1|}=frac{1}{2}ln{|x-1|}-frac{1}{2}ln{|x+1|}+cqquadtext{is OK.}$$

$y(-3)=1 implies c=frac12ln{sqrt{2}}quad$ but not $c=ln{sqrt{2}}$ .

$$2ln{|y+1|}=ln{|x-1|}-ln{|x+1|}+ln(2)$$ $$(y+1)^2=2left|frac{x-1}{x+1}right|$$ Then one have to study the function $f(x)=frac{x-1}{x+1}$ which must be positive. We show that : $$f(x)>0 quadtext{if}quad left(x<0quadtext{or}quad x>2 right)$$ $$f(x)<0 quadtext{if}quad 0<x<2$$ Thus $$(y+1)^2=2frac{x-1}{x+1}quadtext{if}quad left(x<0quadtext{or}quad x>2 right) $$ $$(y+1)^2=-2frac{x-1}{x+1}quadtext{if}quad 0<x<2 $$ The specified point $(x=-3:;:y=1)$ belongs to the branche : $$(y+1)^2(x+1)=2(x-1)qquad x<0.$$

Differential equation, modulus signs in solution?

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