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Let's take the simple example of a logistic differential equation:

$$\frac{dy}{dx} = y(50-y)$$

We get that the solution would be:

$$y(x) = \frac{50e^{50x}}{c_1+e^{50x}}$$

Everything makes sense to me when $0 < y < 50$, it's when we go out of that range that I don't understand how $y(x)$ can be a solution. When $y = 50.1$, for example, the differential equation says that the slope should be $-5.01$. But there is no x-coordinate for $y= 50.1$. Same goes for any y-value equal to or less than zero. So how is this a solution to the logistic differential equation if there are horizontal asymptotes? Shouldn't the differential equation be restricted on its domain to $0 < y < 50$? I am very new to the concepts of differential equations, so the idea of a slope existing without an x-coordinate existing does not make sense to me.

I'll clarify a bit.

Take the point $y=50$. The differential equation says that at $y=50$, the slope of the tangent line would be 0 when the function's y-coordinate is 50. But the issue is that the function will never be 50 or greater, the y coordinate will never exceed or equal 50. So shouldn't the domain of the differential equation be restricted to reflect this fact - that the solution of the diff-eq has a range that is not all real numbers.

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    I didn't get this part: _Shouldn't the differential equation be restricted on its domain to 0 – 2017-02-25
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    Hi- I have tried looking at the slope field, but here's the main question: if I look at y coordinates below 0 or above 50, the slope is negative. How could the slope be negative if nowhere from for any real number x, the slope of y(x) is greater than 0? – 2017-02-25
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    In other words, the diff-eq is giving a tangent line slope for a hypothetical coordinate pair that does not exist, as it's out of range of the solution function. And I don't see why, therefore, there isn't a domain restriction on the differential equation, limiting it to $0 – 2017-02-25
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    "The" solution is not a complete solution. You make some assumptions while developing that family of solutions, and it works for the parts of the domain where those assumptions hold. Obviously $y=0$ is another solution, but you're not going to find it with a solution method that assumes $y\neq 0.$ – 2017-02-25

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You can write this as

$$ y(x) = \frac{50}{Ce^{-50x}+1} $$

Note that, $y(0) = \frac{50}{C+1}$. So if $y(0) = 50.1$ like you picked, $C = \frac{50}{50.1}-1$; i.e., $C < 0 $. So that equation still solves the ODE, but $C$ is different.

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    Thank you Derek, for your answer. I think I may have been unclear - I wasn't talking about what happened at $y(0)$, instead I meant what happens when the y-coordinate of the logistic solution equals zero, which obviously never happens. There is an asymptote there. So how can the differential equation have a solution for when $y=0$ if $y$ never actually *reaches* 0? – 2017-02-25
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    When $y = 0$, C --> infinity, yes. But this makes perfect sense. When $y=0$, $y'=0$ so it never changes, that is, $y$ is identically 0. – 2017-02-25
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    If $y \neq 0$, you're right, the solution will never reach zero in finite time. But if it starts at zero, it stays there and that is still a solution to the ODE. – 2017-02-25
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    Okay, but then how is it that the differential equation provides a $y'$ value for $y < 0$? It does not make sense. – 2017-02-25
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    It depends where you start. If you start at $y=-0.1$, you can still solve for C and that will be your solution. It's all because the C value is arbitrary and depends only on where you want your solution to begin. – 2017-02-25
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    Thank you Derek - how would I start at a negative y value, if the corresponding x value would be negative infinity? – 2017-02-25
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    @rb612: You can't. But on the other hand, you can't start at a positive value either, since all solutions tend to zero as $x \to -\infty$. To get a meaningful initial value problem for this differential equation, you have to pick a finite value $x_0 \in \mathbf{R}$ and start from there, not from $-\infty$. – 2017-02-25
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    @HansLundmark I think he (and I) mean when $x=0$, what can $y$ be... And $y$ can certainly be negative when $x=0$. For example, if you want $y=-1$, then $C=-51$ and that's your solution. – 2017-02-25
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    The graph for an initial condition of $0 < y_{0} < 50$ is a logistic curve with asymptotes at $y=0$ and $y=50$, I think you know that. But for an initial condition elsewhere, the graph dramatically changes. I suggest plotting this solution and varying $c_{1}$ to look at what happens to the solution (desmos.com does a good job of this). – 2017-02-25
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    @DerekOrr: That's how I interpreted the original question too, and you have explained everything perfectly in your answer. But I was referring to rb612's comment just above mine, which I interpreted as “OK, but here's a different question: what's the solution of the ODE if your requirement is $\lim_{x \to -\infty} y(x) = y_0 < 0$ instead of $y(0)=y_0<0$?” – 2017-02-25
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Depending on the value of $c_1$ you can have any solution. For example, for $y=50.1$ we have the solution $x=0$ if $c_1=-0.1/50.1$ $$50.1=\frac{50 e^{50\cdot0}}{c_1+e^{50\cdot 0}}$$ $$50.1 c_1+50.1 =50$$ $$c_1=-0.1/50.1$$

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    You lost a crucial minus sign. – 2017-02-25
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    @Andrei - thanks for your answer. However, I'm not talking about when $x=0$, I'm saying when $y=0$, which for any $c$ will not happen. – 2017-02-25
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    @rb612 $y(x)=0$ or $y(x)=50$ are also solutions. Think about how you got your solution: you divided by $y(y-50)$, which is you assumed $y\ne0$ and $y\ne 50$ – 2017-02-25
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    Great point! That solves the first part of my dilemma. Now how about the values where $y>50$ or $y<0$, those don't exist in the range of the solution function, so what solutions account for this? – 2017-02-25
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    @rb612: Why do you say values $y>50$ aren't covered by your formula, when the answer shows you which $c_1$ to take in order to get $y=50.1$? – 2017-02-25