What am I doing when I separate the variables of a differential equation?

Question

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?

Answer 1

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.

Answer 2

"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.

Answer 3

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.

Answer 4

I encountered a similar problem: I had to study the solution to second order autonomous differential equations. I found it fruitful to first think of symbols like "$dx$" or "$dt$" as small quantities and then take limits. Possibly this is the way Newton and Leibniz thought about infinitesimals.

So, going back to your (very nice) example, let us think about the two integrals as limits of sums. Each summand on the right is of the shape $$ e^{x_k} (x_k - x_{k-1}) $$ (here I've put the function value to the right of the interval, but it doesn't matter, because the exponential function is continuous and the intervals get smaller and smaller in the limit which we are going to take), and on the left hand side there are summands $$ y(x_k) (y(x_k) - y(x_{k-1})). $$

Now we make the intervals smaller and smaller, and due to the continuity of $y$ and the fact that we may set $z_k = y(x_k)$ to simplify the right hand side summands to $$ z_k (z_k - z_{k-1}) $$ (which corresponds to the integral of the function $f(w) = w$), we can already sort of see that we get the right answer. (Of course, we then also have to take the limits of the integration into account; notationally, I've found that it helps using capital letters for these.)

Answer 5

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.