Differential equation, modulus signs in solution?

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.

Mathematics Asked by Refnom95 on December 3, 2021

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 $$y(x)=-1+sqrt{2,frac{x-1}{x+1}}.$$

Answered by Lutz Lehmann on December 3, 2021

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

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

$$y(-3)=1 implies c=frac12ln{sqrt{2}}quad$$ 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 : $$f(x)>0 quadtext{if}quad left(x<0quadtext{or}quad x>2 right)$$ $$f(x)<0 quadtext{if}quad 0 Thus $$(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 The specified point $$(x=-3:;:y=1)$$ belongs to the branche : $$(y+1)^2(x+1)=2(x-1)qquad x<0.$$

Answered by JJacquelin on December 3, 2021

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