On which interval in x is the solution defined? Here is the differential equation.
$$ y' = \frac {2x}{1+2y} $$
Given
$$ y(2) = 0 $$
So as I am solving this I get to 
$$ y+y^2 = x^2-4 $$
I am not sure what the intuition is behind using the quadratic formula at this step...
Can someone please explain why?
 A: The quadratic formula will do two things. First, it'll get you an equation in terms of $y$, instead of $y+y^2$. In the world of functions, this is very ideal. Second, you'll get a radical from the quadratic formula, and you can use it to find an interval for which $y$ is defined on. Remember, (on the real number line) the square root operator is only defined to work on quantities that are greater than or equal to zero.Now you want to solve for $y$ using the quadratic formula, so set the equation equal to zero: $$y^2+y-x^2+4=0 \\ \implies y = \frac{-1\pm \sqrt{1-4\cdot [-x^2+4]}}{2} \\ = \frac{-1\pm \sqrt{4x^2-15}}{2}$$ Again, the quantity under the root must be greater than or equal to zero, so we require $$4x^2-15 \geq 0 \\ \implies 4x^2 \geq 15 \\ \implies |x| \geq \frac{\sqrt{15}}{2}$$ The equation with absolute values tells us we need either $x \geq \frac{\sqrt{15}}{2}$ or $x \leq -\frac{\sqrt{15}}{2}$. The way to represent these inequalities with intervals is to say $x \in \left(-\infty, -\frac{\sqrt{15}}{2}\right] \cup \left[\frac{\sqrt{15}}{2},\infty \right)$.
As @Alan pointed out, the initial condition starts at $x=2$, so we just keep positive the interval to guarantee $y$ has a unique solution.
