TITLE: Quadratic formula in differential equations
QUESTION [9 upvotes]: $$(y')^2 + y' =\frac{y}{x} \tag{0}$$

The solution of this differential equation involves using the quadratic formula for a quadratic in terms of $y'$ but I'm a bit bothered that we get a $ \pm$ when we do that:
$$y'  =- \frac12 \pm \sqrt{\frac{4y}{x} +1} \tag{1}$$
And then we could do $y=xt$ and solve but how exactly do we understand the plus or minus quantity which we get in step-1? It seems that the procedure of completing the quadratic formula generates two differential equation which solves the one in (0). So, should I solve both ones and the actual solution for (0) is a linear combination of both?

REPLY [5 votes]: Expanding upon the comments by @Lutz Lehmann. Given
$$\tag{1}
y'^2+y'=\frac{y}{x}
$$
Let us denote
$$\tag{2}
y'_{\pm}=-\frac{1}{2}\pm\frac{1}{2}\sqrt{\frac{4y_{\pm}}{x}+1}
$$
These are two separate differential equations. For a given initial condition, solving (2) for $y_{\pm}$ yields two permissible solutions to (1). The solution of (1) will not be a linear combination of solutions to (2). Demanding that $y$ is real shows there is a region $y<-x/4$ in which there are no solutions to (1) or (2), which also restricts the initial condition.
A plot is helpful:

The arrows are the vector fields $(1,y'_{\pm})$, plotted as a function of $x$ and $y$ using eq. (2). You could arbitrarily decide that from eg. $1<x<2$, your chosen solution evolves according to $y_+$, then from $2<x<3$ it evolves according to $y_-$, and so on. However, doing so would render your solution non-differentiable at those 'crossover points'$^\dagger$.
Near the critical line $y=-x/4$, the square root $\to 0$ in eq. (2) and we have $y_+' \sim y_-'\sim-1/2$, so the vector field is approximately $(1,-1/2)$. The direction of the critical line is $(1,-1/4)$ so both solutions $y_{\pm}$ (can) flow towards/ past the critical line.
For a given initial condition, the two solutions to (1) are:

The solution $y_+$. For certain initial conditions $y(x_0)=y_0$ this will exist for all $x>x_0$, otherwise it meets the critical line at some $x^*$ and ceases to exist for $x>x^*$.
The solution $y_-$ until it meets the critical line at some $x^*$.

$\dagger$ If it were the case that eg. $y_-$ flows towards but $y_+$ flows away from the critical line, then we could patch together solutions there, and the patched solution would still have a continuous derivative, since $y'_+=y'_-$ on the critical line.