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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?

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4 Answers 4

up vote 16 down vote accepted

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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2  
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. –  the L Mar 16 '11 at 18:59
    
@theL Can you give a reference/more details on this please? –  Lepanais 2 days ago

"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.

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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.

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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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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". –  Mark Eichenlaub Mar 16 '11 at 17:20
    
They can be defined differently as I described. For a more detailed discussion, see en.wikipedia.org/wiki/Differential_of_a_function#Definition. –  Alex Becker Mar 16 '11 at 17:25
    
Thanks for the link, but the description there is not what you said. Differentials as described in Wikipedia are not infinitesimal changes. –  Mark Eichenlaub Mar 16 '11 at 17:30
    
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. –  Alex Becker Mar 16 '11 at 17:37
    
That's naive unless you're willing to work in a non-standard setting, which is far from what the OP wanted. –  Alexei Averchenko Mar 17 '11 at 11:57

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