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?

• Finding a solution here means finding a solution in $y$, so since both $-2C$ and $-x^2$ are degree zero with respect to $y$, we can view them as constants in the above quadratic equation. When we use the quadratic formula to solve $y^2 + 4y -2C - x^2=0$ for $y$, we just lump $8$ and $4C$ into the constant $E$, and keep track of the $x^2$ under the square root. But remember that our final answer is a description of $y$. Jan 21 at 5:47

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.
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 $$y = -2$$): $$y = -2 \pm \sqrt{x^2 + E}.$$