Discarding Plus/Minus Sign

From Elongar's solution at http://www.thestudentroom.co.uk/showpost.php?p=18170264&postcount=21

Question 6, STEP III, 2002

$\displaystyle y^4 \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^4 = (y^2 - 1)^2$

Standard seperating of variables yields:

$\displaystyle \frac{y^4}{(y^2 - 1)^2} (\mathrm{d}y)^4 = 1 . (\mathrm{d}x)^4$

At this point, we would like to take roots, but notice that we must consider seperately the cases $| y | \le 1$, and $| y | > 1$, or we will be taking square roots of a negative number.

(a) For $| y | > 1$:

$\displaystyle \int \frac{y}{\sqrt{y^2 - 1}} \mathrm{d}y = \int \pm 1 . \mathrm{d}x$

Resulting in:

$\displaystyle \sqrt{y^2 - 1} = \pm x + c$

$\displaystyle y^2 = (\pm x + c)^2 + 1$ (?)

Notice that if $x \in [- \infty, \infty]$, it is no longer necessary to consider $\pm x$, and we have:

$\displaystyle \boxed{y^2 = (x + c)^2 + 1}$