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I saw this notation many times, but I don't understand why the $y$ variable is missing in the first term of the first equation below.

$$ \frac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y) $$

It just mean:

$$ \frac{\mathrm{d}y(x,y)}{\mathrm{d}x} = f(x,y) $$

?


Tell me if I'm wrong

In my understanding this is a linear ODE of first order in which $y$ is the unknown function and both of them (the derivate and the unknown one) have just a single variable.

$$ y'(x) = f(x) $$

A simple example

$$ \begin{align} y'(x) &= 2x\\ \int y'(x)\!\mathop{}\mathrm{d}x &= \int 2x\!\mathop{}\mathrm{d}x\\ y(x)&= x^2+\mathrm{C}\\ \end{align} $$

Is this case:

$$ \dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x)) $$

would it be something like this?

$$ \begin{align} y'(x,y) &= 2x + 2y\\ \iint y'(x,y)\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y &= \iint 2x + 2y\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y\\ y(x,y) &= x^2 + y^2+\mathrm{C}\\ \end{align} $$


For Christian Blatter

This is what I would expected to be your exaplanation. Can you tell me why my version is wrong?

When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a slope assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.

Given this "slope field" we are interested in curves $$\gamma:\ x\mapsto \gamma(x)=\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)$$ lying in $\Omega$ that have for each of their points $\bigl(x,\gamma(x)\bigr)$ the slope $f\bigl(x,\gamma(x)\bigr)$ prescribed there. This means that we should have $$\gamma'(x)\equiv f\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$

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    $\begingroup$ In the context of ODEs, most likely $$\frac{d y(x)}{dx} = f\bigl(x, y(x)\bigr).$$ $\endgroup$ Mar 5, 2014 at 18:25

2 Answers 2

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Your difficulty stems from the use of the letter $y$ for two different purposes: (a) as coordinate variable in the $(x,y)$-plane, and (b) as variable for (unknown) functions $x\mapsto y(x)$ whose graphs are lying in the $(x,y)$-plane.

When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a slope assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.

Given this "slope field" we are interested in curves $$\gamma:\ y=\phi(x)\qquad(a<x<b)$$ lying in $\Omega$ that have at each of their points $\bigl(x,\phi(x)\bigr)$ the slope $f\bigl(x,\phi(x))$ prescribed there. Note that these curves $\gamma$ are considered as graphs of unknown functions $\phi:\>x\mapsto y=\phi(x)$. The condition about the slopes means that we should have $$\phi'(x)\equiv f\bigl(x,\phi(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$ Now since centuries it is common to denote the unknown functions in such a problem not by $\phi$, or some other letter, but by the same letter as "the coordinate in which they take values", i.e., $y$ in our case. In view of $(1)$ this means that we are looking for functions $x\mapsto y(x)$ such that $$y'(x)\equiv f\bigl(x,y(x)\bigr)\qquad(a<x<b)\ .\tag{2}$$ The condition $(2)$ is then further abbreviated to $$y'=f(x,y)\qquad\bigl((x,y)\in\Omega\bigr)\ .\tag{3}$$ The ODE $(3)$ encodes the ideas described here and condenses them into four letters plus some extra tokens.

Update concerning your "understanding": The unknown functions are functions of one variable $x$, whether one writes them as $\phi:\>x\mapsto y=\phi(x)$ or as $x\mapsto y(x)$. These functions have derivatives with respect to $x$ that are again functions of the one variable $x$. Therefore expressions of the form $y(x,y)$ or ${dy(x,y)\over dx}$ don't make sense. On the other hand the so-called right side of the ODE, a given function $f(x,y)$ defining the desired slope at each point $(x,y)\in\Omega$ is (in general) a function of the two variables $x$ and $y$. In some special cases, as with the super-simple ODE $y'=f(x)$, or with the ODE $y'=g(y)$ only one variable appears in $f$. The interpretation of this phenomenon is the following: In the first case the assigned slope is constant along vertical lines, and in the second case it is constant along horizontal lines.

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  • $\begingroup$ You got exactly my problem but there are many things which I don't understand in your explanation! Can you take a look at what I've added to my question? $\endgroup$
    – Aurelius
    Mar 6, 2014 at 11:04
  • $\begingroup$ Much better now. Two questions. 1) Why we should use two letters to indicate the curves $\gamma$ and the graph of these curvers $x\mapsto\phi(x)$? 2) Can you make me an example of the "super-simples" ODEs $y'=f(x)$ and $y'=g(y)$? Just to see if I've understood! $\endgroup$
    – Aurelius
    Mar 6, 2014 at 16:10
  • $\begingroup$ ad 1): A curve is a subset of ${\mathbb R}^2$ having a parametrization $t\mapsto\bigl(x(t),y(t)\bigr)$, in some cases $x\mapsto\bigl(x,\phi(x)\bigr)$. A function $\phi:\>x\mapsto y=\phi(x)$ is a map from some $x$-interval to some $y$-interval. – ad 2): The ODE $y'=f(x):=4x^3-\cos x$ has the functions $y(x)=x^4-\sin x + C$ as solutions. The ODE $y'=g(y):=-3y$ has the functions $y(x)=C e^{-3x}$ as solutions. – That's my last word in this matter. $\endgroup$ Mar 6, 2014 at 16:40
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The notation $\dfrac{\mathrm dy(x)}{\mathrm dx}$ is short for $\dfrac{\mathrm dy}{\mathrm dx}(x)$ or $y'(x)$, if you prefer.

In this context, the equality $\dfrac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y)$ should be read as $\dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x))$.

As for your last example, you got the wrong idea, $y$ is a function whose domain is a subset of $\mathbb R$, so $y(x,y)$ doesn't make sense.

Correct would be to consider the function $f\colon \mathbb R^2\to \mathbb R, (x,y)\mapsto 2x+2y$ and to consider the ODE $y'(x)=f(x,y(x))$. Note that in $f\colon \mathbb R^2\to \mathbb R, (x,\color{red} y)\mapsto 2x+2\color{red} y$ the variable $\color{red} y$ is bounded, it's not a function, it's a variable. If you wish you can rewrite the function as $f\colon \mathbb R^2\to \mathbb R, (x,u)\mapsto 2x+2u$ and then you can consider the ODE $y'(x)=f(x,y(x))$ without any ambiguity.

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  • $\begingroup$ I'm used to see the derivative operator as $\dfrac{\mathrm d}{\mathrm dx}[y(x)]$ or $\dfrac{\mathrm dy(x)}{\mathrm dx}$ has your notation the same meaning? Can you make me a very simple example with a real ODE? $\endgroup$
    – Aurelius
    Mar 5, 2014 at 18:36
  • $\begingroup$ The derivative operator is a function that takes as input a function. The symbol $y(x)$ is not a function (usually). The symbol $\dfrac{\mathrm d}{\mathrm dx}[y(x)]$ is an abuse of notation to, perhaps, make clear what the variable is. if you want to look at it as an operator, correct would be $\dfrac{\mathrm d}{\mathrm dx}(y)$ where $y$ is a differentiable function. Do you still require an example? Of what exactly? $\endgroup$
    – Git Gud
    Mar 5, 2014 at 18:51
  • $\begingroup$ If $y(x)$ is not a function usually, what would it be? For the example I'll let you know! $\endgroup$
    – Aurelius
    Mar 5, 2014 at 19:33
  • $\begingroup$ @FormlessCloud $y(x)$ is the value of $y$ evaluated at $x$. For instance $2x$ isn't a function. The intended function is $\colon \mathbb R\to \mathbb R, x\mapsto 2x$. $\endgroup$
    – Git Gud
    Mar 5, 2014 at 19:36
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    $\begingroup$ @FormlessCloud I didn't. That's why I said I was lazy $\ddot\smile$ Anyway, is there anything I can make more clear? I'm not exactly sure I helped you. $\endgroup$
    – Git Gud
    Mar 5, 2014 at 19:50

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