# Quadratic formula in differential equations

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

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?

• I think you can split yur problem into two cases. And solve both cases independently- Dec 3, 2020 at 11:23
• @TitoEliatron my main problem is interpretting and secondly would the actual diff eqn's soln be linear combination of both solns? Dec 3, 2020 at 11:25
• Yes, you solve both and then you will have to paste the solutions together. Each time $y'$ changes sign, you pass from one solution to the other. Dec 3, 2020 at 11:26
• I didn't understand how the changing sign of y' suggests changing solutions from one to another. Could you rephrase that? Dec 3, 2020 at 11:27
• This was not correct. Each time $y'$ has a double root, that is, if $y'=-\frac12$, you can switch branches of the root without losing the defining characteristic that an ODE solution should be continuously differentiable. The equation is not linear, thus there is nothing to do with linear combinations. One could check if the term under the root defines an extra solution, $y=-\frac x4$, which appears to be not the case. Dec 3, 2020 at 12:04

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.

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. $$1, your chosen solution evolves according to $$y_+$$, then from $$2 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.

• How did you get this vector field? Sep 27, 2021 at 3:19
• Intuitively: pick a point $y_0(x_0)$, which would be at the co-ordinates $(x_0,y_0)$. If we increase $x_0$ by some amount, say $\delta x$, this point flows to the point $(x_0 + \delta x, y_0 + y' \delta x)=(x_0,y_0)+(\delta x, y' \delta x)$. So the direction of the flow is $\delta x(1,y')$
– Sal
Sep 27, 2021 at 3:43

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 $$C=1,2,3$$

• In differentiate with y step, why were you able to treat x as constant Dec 3, 2020 at 13:05
• @Lutz Lehmann Thanks you may see my solution now. Dec 3, 2020 at 15:32
• @Buraian Thanks you may see my solution now. Dec 3, 2020 at 15:33
• @Lutz Lehmann Thanks I have corrected it now. Dec 3, 2020 at 15:54
• I still don't get how you differentiated with 'y', also I think you are missing an integral sign in that step Dec 3, 2020 at 17:35