# How to show that the positive x-axis consists of orbits for a system of differential equations?

I have a homework question that I'm working on, and I'm trying to make sure I understand how orbits work with regards to solutions of differential equations.

I have a system of differential equations as so:

$$x' = a - x -\frac{6x}{(3+x)}y \\y' = (\frac{6x}{(3+x)}-1)y$$

Where $x'$ is the derivative of $x$ with respect to $t$ and $y'$ is the derivative of $y$ with respect to $t$.

These are meant to represent a biological growth problem, so we're just analyzing the behavior in quadrant I.

The question is to show that the positive x-axis consists of orbits.

My understanding is that the orbits are the curves generated by the set of points $$(x(t), y(t))$$ associated with the general solution of the system, so since I'm looking to show that the x-axis consists of orbits, I want to find the curve generated by the set of points $$(x(t),0)$$

I am not completely sure of the right way to do this. What I did (And what I'm asking for verification on) is the following:

$$y(t) = 0 \implies \\ x'=a-x \\ y' = 0$$ Then I solved that system and got: $$x(t) = a - ce^{-t}$$

As per dbx's comment, I found a value for $c$ by solving for $x(t_0)=x_0$, from which I got $c = a - x_0$, so my general solution is:

$$x(t) = a - (a-x_0)e^{-t}$$

I then planned to substitute $x(t) = a - (a-x_0)e^{-t}$ and $y(t)=0$ into my original system to show that this solution $(x(t),y(t))$ is a valid solution, however it's based on my assumption that I can just zero out the $y$'s in my original $x'(t), y'(t)$ system.

Is it correct to do that? And if so, is my final answer $x(t) = a - (a-x_0)e^{-t}, y(t) = 0$ correct to show that the positive x-axis consists of orbits?

• Is solution-verification not the right tag for this? It seems to be synonymed with proof-verification ; even when I try retagging it as solution-verification it goes back to proof-verification, though since this isn't a proof I thought the other tag would have been right. EDIT: I found This meta question That explains why the solution-verification tag is worthless and shouldn't be used. Dec 5, 2017 at 13:20
• Think in terms of the DEs that describe the system. If you have an initial condition $(x_0, 0)$ on the $x$-axis, what will happen to it? That is, what will be $x’$ and $y’$? Dec 5, 2017 at 13:28
• @dbx In that case I'd get $x_0 = a - ce^0$ which simplifies to $c = a - x_0$ -- I didn't think about that, and that lets me get rid of my invented constant $c$, thanks! Is there anything else you see? Dec 5, 2017 at 13:33
• Where is the parameter t present in your DE? Dec 9, 2017 at 19:53
• @YourIDE Sorry I didn't clarify that, all these $x'$ and $y'$ are ${dx}/{dt}$ and ${dy}/{dt}$. I got too overly used to my teacher's notation that I forgot that outside of my class, usually prime meant with respect to that same variable, even though our textbook uses ' explicitly for derivative with respect to $t$. I've updated the question. Dec 9, 2017 at 20:00