I see an equation like this:

$$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$

and solve it by "separating variables" like this:

$$y\textrm{d}y = e^x\textrm{d}x$$ $$\int y\textrm{d}y = \int e^x\textrm{d}x$$ $$y^2/2 = e^x + c$$

What am I doing when I solve an equation this way? Because $\textrm{d}y/\textrm{d}x$ actually means

$$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$

they are not really separate entities I can multiply around algebraically.

I can check the solution when I'm done this procedure, and I've never run into problems with it. Nonetheless, what is the justification behind it?

What I thought of to do in this particular case is write

$$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$ $$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$

then by the fundamental theorem of calculus

$$y^2/2 = e^x + c$$

Is this correct? Will such a procedure work every time I can find a way to separate variables?

  • $\begingroup$ Please see this for why we cannot anyhow solve so-called separable differential equations blindly (and hence why your question is a good one). See my comments starting here for extra details. $\endgroup$ – user21820 Mar 24 '19 at 9:00
  • $\begingroup$ Related to that, the answer by Mike Spivey is wrong, because it fails to get the right answer to the differential equation in the linked post. $\endgroup$ – user21820 Mar 25 '19 at 8:06

The basic justification is that integration by substitution works, which in turn is justified by the chain rule and the fundamental theorem of calculus.

More specifically, suppose you have: $$\frac{dy}{dx} = g(x) h(y)$$ Rewrite as: $$\frac{1}{h(y)} \frac{dy}{dx} = g(x)$$ Add the implicit dependency of $y$ on $x$ to obtain $$\frac{1}{h(y(x))} \frac{dy}{dx} = g(x)$$

Now, integrate both sides with respect to $x$: $$\int \frac{1}{h(y(x))} \frac{dy}{dx} \, dx = \int g(x) \, dx$$ If we do a variable substitution of $y$ for $x$ on the left-hand side (i.e., use the integration by substitution technique), we replace $\frac{dy}{dx} dx$ with $dy$. Thus we have $$\int \frac{1}{h(y)}\, dy = \int g(x) \, dx,$$ which is the separation of variables formula.

So if you believe integration by substitution, then separation of variables is valid.

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    $\begingroup$ you can also give it a formal justification by working with differential forms. The only difficulty is why "dy/dx" is defined, and I think the answer is that under reasonable assumption, the module of 1-forms is free of rank 1, so you may define "divison" of two forms and the answer is a function. $\endgroup$ – the L Mar 16 '11 at 18:59
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    $\begingroup$ @theL Can you give a reference/more details on this please? $\endgroup$ – Anthony Conway Aug 24 '14 at 22:07
  • $\begingroup$ @Mike spivey Is there any chance you could write the substitution explicitly? In specific what is the "u" here? I am pondering about this as treating the dxs and dys as ratios won't work if it's attempted as a solution for a differential equation in standard form (ie. not dividing by 1/y). $\endgroup$ – Dole Jun 28 '16 at 19:02
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    $\begingroup$ @Dole: I'm using $y$ instead of $u$. A variable switch from $x$ to $u$ would have $du = \frac{du}{dx} dx$ for the $dx$ switch to $du$. So I'm just using $y$ instead of $u$. $\endgroup$ – Mike Spivey Jun 28 '16 at 21:25
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    $\begingroup$ @MrReality: At some places in the derivation I wanted to emphasize the dependence of $y$ on $x$, and at other places I wanted to de-emphasize that dependency. $\endgroup$ – Mike Spivey Oct 27 '17 at 17:49

"Separation of variables" in ODE (which has nothing to do with separation of variables in PDE) is a kind of magic that is easy to perform but difficult to justify.

Assume that in the given differential equation the quantities $x$ and $y$ are functions of a hidden variable $t$ (time). Then the equation $y\>y'=e^x$ is equivalent to $y(t){\dot y(t)\over \dot x(t)}\equiv e^{x(t)}$, resp. $$y(t)\dot y(t)\equiv e^{x(t)}\dot x(t).$$ Integrating this from $t=0$ to $t=T$ one gets $${1\over2}(y^2(T)-y_0^2)=e^{x(T)}-e^{x_0},$$ where $(x_0,y_0)$ is the initial condition and $T$ is arbitrary. This means: At any given time the quantities $x$ and $y$ are related by the equation $${1\over2}(y^2-y_0^2)=e^x-e^{x_0}.$$ Looking back, one can see that the relation between $x$ and $y$ obtained in this way is exactly the equation obtained by following the recipe given in the books.


maybe its better to think of it as $y\frac{dy}{dx}=e^x$. the two functions of $x$ are equal, so their indefinite integrals (with respect to $x$) are equal (i.e. the way you talked about it at the end). moving the "differentials" around is more of a convenience.


Seperation of variables involves manipulating the differentials (the $dx$'s and $dy$'s in your equation). A differential is the infinitesimal change in a variable, and can be treated as a variable in its own right in many applications. With this perspective, $dy$ is a function of $x$ and $dx$, and the derivative $dy/dx$ is the ratio of these two differentials, which is a function of $x$. What you are doing is simply performing an algebraic manipulation of these variables and then using calculus to remove the differential terms.

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    $\begingroup$ The $dx$ and $dy$ are coming from a single limit. That's what I've been taught in calculus. This answer is logically equivalent to saying "you can do that because you can do that". $\endgroup$ – Mark Eichenlaub Mar 16 '11 at 17:20
  • $\begingroup$ They can be defined differently as I described. For a more detailed discussion, see en.wikipedia.org/wiki/Differential_of_a_function#Definition. $\endgroup$ – Alex Becker Mar 16 '11 at 17:25
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    $\begingroup$ Thanks for the link, but the description there is not what you said. Differentials as described in Wikipedia are not infinitesimal changes. $\endgroup$ – Mark Eichenlaub Mar 16 '11 at 17:30
  • $\begingroup$ Yes, they are. Infinitesimal change is the change in the linear approximation, which is what is being used in that definition if you study it closely. $\endgroup$ – Alex Becker Mar 16 '11 at 17:37
  • $\begingroup$ That's naive unless you're willing to work in a non-standard setting, which is far from what the OP wanted. $\endgroup$ – Alexei Averchenko Mar 17 '11 at 11:57

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