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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?

4 Answers 4

43

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.

  • 1
    @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.2017-10-27
10

"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.

  • 0
    Leibniz's calculus of differentials isn't exactly radical.2011-03-17