Why is this instance correct of having a plus or minus sign, outside of a square root? I was really struggling with a differential equations question, when I came across the solution on this website. I understood everything of the solution, but I did not understand the final part (where the solver added a $\pm$ sign to the right side of the equation). The original problem is to find a non-constant solution to ${x'}^2+x^2=9.$
I subtracted $x^2$ from both sides, leaving ${x'}^2=9-x^2.$
Then I took the square root of both sides (is this where it gets messy?), which left me with $x'=\pm\sqrt{9-x^2}.$
Then, I divided $\sqrt{9-x^2}$ by both sides:  $$\frac{x'}{\sqrt{9-x^2}}=1.$$
Afterwards, I took the integral of both sides and rewrote $x'$ as $$\int \frac{x'}{\sqrt{9-x^2}} dx= \int 1 dt.$$ 
Using the formula integration guy for arcsin, I got: $$\arcsin\left(\frac x3\right)+c_1=t+c_2. 
$$
Then I took the sin of both sides and simplified the arbitrary constants: $x/3=\sin(t+c_3).$
And then I multiplied by $3,$ leaving us with $x=3\sin(t+c).$ 
I will stop there for now, as the rest (er initial condition(s)) do not pertain to my question. 
When I was reading the solution, the solution contained a $\pm$ on the right hand side of the solution at the end. Can someone explain why to me please? I don't quite understand...
It would be greatly appreciated!
 A: In general, $\sqrt{x^2}$ is not equal to $x$, it is equal to $|x|$, the absolute value of $x$. To see this, think about what happens when you calculate something like $\sqrt{(-1)^2}$: the answer is not $-1$, it is $|-1|=1$, because the square root is always nonnegative when defined.
So after your first step, you really get:
$$\begin{align*}
(x’)^2 &= 9-x^2\\
\sqrt{(x’)^2} &= \sqrt{9-x^2}\\
|x’| &= \sqrt{9-x^2}.
\end{align*}$$
Now, there are two ways in which you can have $|x’|=\sqrt{9-x^2}$: you can have either $x’ = \sqrt{9-x^2}$, or you can have $x’ = -\sqrt{9-x^2}$.
You can try working each of them separately, or keep track of both possibilities together using a $\pm$. 
Now, your next step has to be done carefully. When you divide by $\sqrt{9-x^2}$, you are implicitly asserting that this quantity is not zero. That is, you are in a sense excluding the possibility that $x=3$ or $x=-3$. Is this a valid exclusion? No; both of these are solutions to the original equation. So you should keep track of these answers that you (inadvertently) dropped.
Then: when you divided by $\sqrt{9-x^2}$ (under the assumption that $x$ is not the constant function $3$ and not the constant function $-3$) you should have kept track of the two possibilities for $x’$: you don’t get $1$ on the right hand side, you get either $1$ or $-1$. That is where your missing $\pm$ comes from. 
(The initial conditions may have allowed you to discard the possibility that $x=3$ or $x=-3$ hold; but since you did not give us the original conditions, and you explicitly say you aren’t worrying about them right now, it is worth mentioning that you need to be careful when you divide by a quantity involving the unknown....)
A: When you divided by $\sqrt{9-x^2}$, the right hand side should have been $\pm 1$ rather than $+1$. Suppose you know that $a = \pm 2b $. What can you say about $a/b$? Well, $a = 4, b = 2$ is one solution, so $a/b$ could be $2$. But $a = 6, b = -3$ is another solution, so $a/b$ could be $-2$. So if $a = \pm 2b$, then $a/b = \pm 2$. Your situation is just a fancier version of this.
