Confused about first order ODE solution using quadratic equation I am working a few problems in my textbook for practice and I came across this problem:
$y' = \frac{x}{y+2}$
Proceeding as usual:
$\frac{dy}{dx} = \frac{x}{y+2}$
Separating the equation:
$(y+2)dy = xdx$
Taking the integral of both sides:
$\frac{1}{2}y^2 + 2y = \frac{1}{2}x^2 + C$
Simplifying slightly:
$y^2 + 4y = x^2 + 2C$
Now, after finding the worked solution for this problem (I was lost at how to resolve this):
$y^2 + 4y - 2C -x^2 = 0$
From this point the solution seems to treat both 2C and x^2 as part of the constant term of a quadratic equation:
$\frac{-4 \pm \sqrt{16 + 8C + 4x^2}}{2}$
$-2 \pm \sqrt{8+4C+x^2}$
Which is then simplified with a variable:
$-2 \pm \sqrt{x^2 + E}$
My question for this is what allowed the two terms (one of them being the independent variable) to be moved over and treated like a constant together with the constant C? Perhaps I am just getting lost in notation but in a quadratic equation:
$Ax^2 + Bx + C = 0$
The C is implied to be just a constant. What am I missing here?
 A: Perhaps thinking of it like this would help. When you try to write $y$ in terms of $x$, you are trying to write a formula for how you would work out $y$ if you knew $x$. For instance if $x$ had the value 3, you would have the equation $y^2+4y-(2C+3^2)=0$ and you would use the quadratic formula to find the possible values of $y$. In the general case, you are simply summarising what you would do if $x$ had any specific value, so you are solving $y^2+4y-(2C+x^2)=0$, thinking about $x$ temporarily as a fixed but unknown number.
A: The method of separation of variables yields an equation of the form
$$
G(y) = F(x) + C, 
$$
for some functions $F$ and $G$ and a constant $C$. There's no reason, in general, that such an implicit equation can be rearranged to express one variable as a function of the other, i.e. of the form $y = f(x)$. Think of the vertical line test: a vertical line of the form $x = a$ will cross the graph of a given solution $G(y) = F(x) + C$ multiple times, namely all the solutions $y$ to $G(y) = F(a) + C$, so at best we can hope to write a family of functions $f_1, \dots, f_n$, the union of whose graphs form the solution curve.
In this example, both $G(y)$ and $F(x)$ are quadratic, so we can get away with two functions if we insist on writing $y = f(x)$. Starting with
$$
y^2 + 4y = x^2 + 2C, 
$$
rather than using the quadratic formula on
$$
y^2 + 4y + (-2C - x^2) = 0, 
$$
it's a bit more straightforward to complete the square in $y$:
$$
y^2 + 4y + 4 = x^2 + 2C + 4, 
$$
which leads to
$$
(y + 2)^2 - x^2 = E, 
$$
where $E = 2C + 4$ is an arbitrary real number. This is easily recognizable as the family of hyperbolas whose asymptotes are the lines $y = \pm x - 2$.

For a particular value of $E$, you can, of course, solve for $y$ to get equations for the upper and lower halves of the hyperbola (on either side of $y = -2$):
$$
y = -2 \pm \sqrt{x^2 + E}. 
$$
