5
$\begingroup$

$\def\d{\mathrm{d}}$I understand what differential equations are and what they are useful for. They're very interesting, but I'm not quite sure what is going on when actually solving one.

This is what I'm trying to solve:

$$\frac{\d y}{\d x} = 2y + 3.$$

As far as I know, this is a first-order linear ordinary differential equation. Seems fairly simple.

The first step that I am told to do is to 'separate the variables'. So I multiply both sides of the equation by $\d x$ and then divide both sides by $2y + 3$ to end up with:

$$\frac{\d y}{2y+3} = \d x.$$

So now I've got all of my $y$ terms on one side, and all of my $x$ terms on the other.

The fact that I can split apart the $\frac{\d y}{\d x}$ does seem a bit strange to me, but I'm told that it's sensible and I'm willing to believe, for now, that that's an OK thing to do.

After this, I'm told to 'integrate both sides' and I'm shown this as the next step:

$$\int \frac{\d y}{2y+3} = \int \d x$$

Normally when integrating, I would denote the 'integral of [integrand] with respect to $x$' using the standard notation:

$$\int \text{[integrand]}\,\d x$$

In definite integration, the $\int$ represents an infinite sum, and the $\d x$ represents, as usual, a very small change in $x$ that is being multiplied by the value of the function at that point in order to, as $\d x$ approaches zero, generate the exact value of the area under the curve.

For now I will just accept that we also write it in this fashion when doing indefinite integration. However, the second step in the solution to the DE confuses me because it doesn't seem to follow this notation.

If you want to integrate both sides, then that means we must be integrating $\d x$ on the right hand side. If this is true, then I would normally write it as $\int \d x\,\d x$, because it is the integral of $\d x$ with respect to $x$. However, in the solution it simply shows $\int \d x$ as if we have lost the $\d x$ that we would have put in as notation.

The other side of the equation also doesn't have a $\d y$ at the end, as I would expect it to.

My real question is: What do $\d x$ and $\d y$ represent on their own? I understand that $\dfrac{\d y}{\d x}$ represents the derivative of $y$ with respect to $x$, and that it is representing an instantaneous rate of change because it is essentially representing an infinitesimally small change in $y$, divided by an infinitesimally small change in $x$ (the gradient at a point).

However, when I see something like $\d x$ on its own, I'm not sure what the meaning of it is any more. It's an infinitesimally small change in $x$, right? How do we integrate that? What does it mean? I want to be very clear that I'm not asking someone to walk me through solving it, because the web is full of examples of solving these types of equations. I'd be much more grateful for a conceptual answer to my question.

And as an extra question: How does integrating both sides of an equation with respect to different variables make sense? Surely the two sides would no longer be equal.

$\endgroup$
  • 2
    $\begingroup$ Good question. You'll find some useful information in these posts: math.stackexchange.com/a/27433/159845, math.stackexchange.com/a/182352/159845 & math.stackexchange.com/a/1252426/159845 $\endgroup$ – StackTD Apr 20 '17 at 9:28
  • 1
    $\begingroup$ Please have a look at this post, of which yours may be a duplicate. $\endgroup$ – Harry49 Apr 20 '17 at 9:29
  • $\begingroup$ @StackTD Thank you, but none of those posts seem to answer my exact question. I'm asking specifically why the usual trailing $dx$ is lost on the right hand side when integrating with respect to $x$. $\endgroup$ – 雨が好きな人 Apr 20 '17 at 17:43
  • $\begingroup$ But it's not missing. The wrangling that you did to get the terms to each side restructured things so you are essentially taking the integral of $\frac{1}{2y+3}$ with respect to y on the left and the integral of 1 with to x on the right. $\endgroup$ – scrappedcola Apr 20 '17 at 19:05
3
$\begingroup$

In this question, we actually don't need to ponder what $\newcommand{\D}{\mathrm{d}} \D y$ and $\D x$ mean individually (in terms of differential forms, etc.). Instead, this splitting apart of $\D y \over \D x$ can be justified in the following more rigorous manner. After separating variables, you have the equation

$$\frac{1}{2y(x) + 3} \frac{\D y}{\D x} = 1 \quad \text{or} \quad \frac{1}{2y(x) + 3} y'(x) = 1 $$ where we should keep in mind that $y$ is a function of $x$ (dependent / independent variables). Now, we integrate both sides with respect to $x$:

$$\int \frac{1}{2y(x) + 3} \frac{\D y}{\D x} ~\D x=\int \frac{1}{2y(x) + 3} y'(x) ~\D x = \int 1~\D x.$$

Observations:

  1. For the right hand side, we recognize that $\int ~\D x = \int 1 ~\D x$. This is where that missing integrand you referred to goes. These truly give the same anti-derivative. In higher math, $\int \D x$ is probably preferred, but this requires training in something like differential geometry / differential forms / tensor calculus / calculus on manifolds.
  2. For the left hand sides, we can recognize this as the integral of a chain rule. Taking the total differential of $y(x)$ gives $$dy = \frac{\D y}{\D x} \D x = y'(x) \D x.$$ Thus the left hand side becomes $$\int \frac{1}{2y + 3} ~\D y.$$ If you prefer, you can think of this a $u$-sub: let $u(x) = y(x) \implies \D u = \D y = \frac{\D y}{\D x} \D x$ and the integral becomes $$\int \frac{1}{2u + 3} ~\D u.$$

What's going on is that the Leibniz notation for calculus is a little too good---many of the standard calculus rules appear to just be manipulating fractions. For example,

  • $u$-sub with Leibniz: $\int f(u(x)) \frac{\D u}{\D x} ~\D x = \int f(u) ~\D u$. This almost looks like we just cancel the "$\D x$".
  • Chain rule with Leibniz: $\frac{\D y}{\D t} = \frac{\D y}{\D x}\frac{\D x}{\D t}$.
  • Total differential with Leibniz: $\D y = \frac{\D y}{\D x} \D x = y'(x) \D x$.
  • Stokes' Theorem: $\int_{\partial M} \omega = \int_M \D \omega$.

Despite how suggestive this notation is, you should always keep in mind that $\frac{\D y}{\D x}$ is not really a fraction. However, it does behave like a fraction in many ways and this can be a useful way to remember these calculus properties.

General Separable Problem

Given a separable differential equation

$$f(y(x)) \frac{\D y}{\D x} = g(x) $$

we integrate both sides with respect to $x$ to get

$$\int f(y(x)) \frac{\D y}{\D x} ~\D x = \int g(x)~ \D x.$$

As above, we recognize the left hand side as the result of a chain rule / a $u$-sub integral to get $$\int f(y) ~\D y = \int g(x) ~\D x.$$

Note that this gets us to the same place as if we had split up $\D y \over \D x$ to begin with: \begin{align*} f(y) ~\frac{\D y}{\D x} &= g(x) \\ f(y) ~\D y &= g(x) ~\D x\\ \int f(y) ~\D y &= \int g(x) ~\D x. \end{align*} While this second version should make a mathematician a little squeamish, it is justified in this case by our reasoning above and is a useful computational shortcut.

Addendum

If you really want to know what $\D x$ and $\D y$ truly mean, this Introduction to Differential Forms gives a good overview, written for an undergraduate audience (multivariable calculus as only prerequisite). The glib answer is that $\D x$ returns the $x$ component of a vector...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.