Quadratic formula in differential equations

Question

$$(y')^2 + y' =\frac{y}{x} \tag{0}$$

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?

Answer 1

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:

1. 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 $x>x^*$.
2. 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.

Answer 2

Incomplete solution:

Let $y'=p$, then the ODE is $$p^2+p=\frac{y}{x} \implies x=\frac{y}{p^2+p}~~~~~(1)$$ D. (1) w.r.t. $y$ and write $\frac{dx}{dy}=\frac{1}{p}$ to get $$p^3+p^2=-y(2p+1)\frac{dp}{dy} \implies \int \frac{dy}{y}=-\int\frac{2p+1}{p^3+p^2}dp$$ $$\ln Cy=\frac{1}{p}-\ln p+\ln(1+p) \implies y=\frac{(1+p)e^{1/p}}{Cp}~~~~(2)$$ Putting this in (1), we get $$x=\frac{ e^{1/p}}{Cp^2}~~~~(3)$$ (2) and (3) give the complete solution of the first order ODE (1) in terms of one undetermined constant $C$, here $p$ acts as a real parameter. One may eliminate $p$ between (2) and (3) to get the cartesian solution with one constant $C$. Se the solution $y(x)$ for $C=1,2,3$