Confused about first order ODE solution using quadratic equation

Question

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?

Answer 1

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.

Answer 2

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