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On the exam I was asked the question about Transcritical bifurcation. I gave the equation

$$ \dot x = rx - x^2 $$

Then I was asked why it is not a differential equation and I couldn't answer. I thought if some derivative equals a function - it is a differential equation.

It's not a differential equation, because r is a variable not x?*

*Sorry If something is not clear. I suffer for lack of math in english.

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    Why not? If $r$ is a parameter, this is a differential equation.2012-03-05
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    Not sure but in bifurcation analysis r is a variable.2012-03-05
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    It is not linear differential equation but it is a differential equation because of x'.2012-03-05

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Definition 1. A differential equation is an equation that involves the derivatives of an unknown function of one or more variables. (Spiegel)

I personally like to change involves by relates.

Since we have an unknown function $x(t)$ and an equation that involves the function $x$ and its derivative:

$$x'(t) = rx(t)-x(t)^2$$

that is a differential equation, and the function you're looking for is

$$x(t) = \frac{r \cdot c \cdot e^{rt}}{1+c\cdot e^{rt}}$$

where $c$ is arbitrary.

I think you might have a bone to pick with your examiner.

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The equation $$\dot{x}(t)=rx(t)-x^2$$ is a family of $ODEs$ because the parameter $r$ can assume infinite values. So it's not an equation, but a family of infinite equations.

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    Correction: $r$ cannot assume infinite values, though it can assume any of infinitely many values. (To say that $r$ can assume infinite values is to say that it can take on values like $\infty$.)2012-03-05
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    I disagree: $\dot{x}(t)=rx(t)-x^2$ without any further qualification **is** a differential equation. To specify the family, one must indicate that one is considering the whole family over some range of values of $r$.2012-03-05
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    @Brian M. Scott: I agree. It depends on my bad english.2012-03-05
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    I figured as much, and I wouldn’t have said anything, except that I’ve heard native speakers make the same mistake (‘infinite values’ instead of ‘infinitely many values’), so I thought that I’d better mention it.2012-03-05
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    In the logistic equation r and x must be positive defined. In the equation above, they can assume all values on the real axis. So the family of ODEs is defined as all the possible differential equation obtained varying the parameter r2012-03-05
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    I'm no expert in differential equations, but to my outsider eye this seems like an unusually pedantic distinction to make. For instance, if I said "the function $f(x) = ax+b$" and someone replied "No, that's not a function. It's a *family* of functions depending upon the parameters $a$ and $b$," I would probably roll my eyes. Is there something going on here that makes this analogy inappropriate?2012-03-05
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    @Pete: I agree completely.2012-03-05
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    @PeteL.Clark I was just going to comment: "Do we call $ax^2+bx+c=0$ a family of equations?"2012-03-06
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    @PeteL.Clark: it's the only way to defend the statement the equation above is not an ODE2012-03-08