When we say something like: $$ \frac{dy}{dx} = x$$ we're describing the way a function $y$ changes with respect to $x$. To solve the differential equation we integrate both sides. Is there a proper way to do this, or is it simply the case that if have a reliable method to get the correct answer then that's fine. For example:

Method 1: $$ \int\frac{dy}{dx} dx = \int x \, dx$$ $$ \int dy= \int x \, dx$$ $$ y + c_1 = \frac{x^2}{2} + c_2$$

Method 2: $$ \frac{dy}{dx} = x$$ $$ dy = x \, dx$$ $$ \int dy= \int x \, dx$$ $$ y + c_1 = \frac{x^2}{2} + c_2$$

This is the thinking that lead me to the question to the title.

  1. Why is it okay to integrate both sides of an equation? Are we basing that on the fact that indefinite integration is simply the reverse of differentiation? Will integrating both sides always make sense with continuous, integrable, elementary functions?

  2. Can we integrate without respect to a variable? In method 2 I employed the integration operator on both sides without adding a differential (I know it was already there).

  3. If I write $ y'' + 5xy = 2x^2$, is it acceptable to integrate both sides of the equation with respect to y, or x, and get a valid equality as a result?

  4. When I write: $ xy + y^x = 2x + 3$, maybe this is bad example, but am I stating that the left side is the same as the right side, or am I proposing that there are certain values of x and y that satisfy that equation.

I'm having sort of a fundamental crisis here. I don't think I was ever properly taught what equality means. If you were, where did you learn it?

  • 2
    $\begingroup$ For starters, you integrated incorrectly. $y' = x$ should lead to $y(x) = \frac{1}{2} x^2 + C$. As for the concern, do you know what it means to take an "integral" $\int$? What sort of definitions were you given? $\endgroup$ – Gregory May 23 '17 at 18:02
  • 1
    $\begingroup$ To your first question, an equality always means that the "thing" on the left hand side is the same "thing" as the "thing" on the right hand side. There is no difference in any respect whatsoever between the two "things". If there is some difference, then they are not equal and the equation is incorrect. Therefore, integrating one side of a correct equation must be equivalent to integrating the other side. $\endgroup$ – bob.sacamento May 23 '17 at 18:07
  • $\begingroup$ @Gregory I'm not sure I know what it means to take an "integral" that's part of the post. I know the definition of the Riemann integral, maybe not as well as you, but the part I don't understand is the equivalence of two riemann integrals, especially when they are indefinite. The only understanding I have is that the indefinite integral of a function is the antiderivative (family of functions with a constant). To take an integral without respect to a variable or with respect to different variables on either side of the equation is what's perplexing me. My elementary calc text doesn't comment. $\endgroup$ – Zduff May 23 '17 at 18:14
  • $\begingroup$ @bob.sacamento That seems like a very unfortunate formulation, since it is rarely obvious from what we write what the "thing" actually is. If I write $\sqrt{2}$ I usually mean $\{x;x^2<2\wedge x\in \mathbb{Q}\}$ but occasionally I also mean $\{\{a_n\}_{n=0}^\infty;\lim_{n\rightarrow\infty}a^2_n=2\wedge \forall n \;a_n\in\mathbb{Q}\}$ and I'd better hope those two are equal for almost all intents. The point I'm trying to make is that euality is an overloaded operator that is generally an equivalence relation regarding the currently chosen base of operations. $\endgroup$ – DRF May 23 '17 at 19:57
  • 1
    $\begingroup$ @DRF Alright, let's chat. How do I open a room? $\endgroup$ – Zduff May 24 '17 at 15:36

I think some of the source of your confusion is thinking about "integrating both sides" and in using variables somewhat indiscriminately. There are ways to do both of those things correctly after you've developed some intuition, but at the beginning of your study you should concentrate more on the meaning.

A better way to ask your question is

The only thing I know about an unknown function $f$ is that $f'(x) = x$, What can I say about $f$?

Well, maybe I'm lucky and can guess one such function: $f(x) = x^2/2$. (That might be more than just luck. I might actually remember something about integration). Then I can invoke a theorem that tells me that any two functions with the same derivative must differ by a constant. That tells me the only possibilities for $f$ are $$ f(x) = x^2/2 + c $$ for some constant $c$.

If I'm not so lucky, and can't guess or figure out a function whose derivative is what I want, I can still answer the question, but in a different way. For example, I might want to find $f$ when I know $$ f'(x) = e^{-x^2} \text{ or } f'(x) = \cos(x + \sin(x)). $$ (The first of these is very useful. The second is just made up ugly. There is no good "formula" for $f$ in either case.)

The way to proceed then is to define $f$ as a definite integral. In the first example you could let $$ f(x) = \int_0^x e^{-t^2} dt $$ and calculate that value for any particular $x$ numerically: it's just a well defined area. The derivative of the function defined by tabulating those values is in fact just the given $f'(x) = e^{-x^2} $. That's (essentially) the fundamental theorem of calculus.

In the other examples you ask about the derivative(s) of the unknown function are mixed up with the function values in much more complicated ways than simply $$ f'(x) = \text{some expression involving just } x. $$ Then you've begin the study of differential equations.

Long winded answer. Hope it helps.


Method $2$ would be considered more appropriate and note $\int x dx = \frac{1}{2}x^2 + C$. However, you only need one arbitrary constant in the solution, $y = \frac{1}{2}x^2 + C$ would suffice for a solution since $c_2 - c_1$ would result in a new constant, where we call it $C$.

1) It is okay to integrate both sides of an equation when we have separated our variables. In other words, when we have $f(x,y)dx + g(x,y)dy = h(x,y)$, we manipulate it to look like $\alpha(x)dx = \beta(y)dy$. However, sometimes this is not feasible so the topic of Ordinary Differential Equations (ODEs) is introduced to some of the cases where it is not separable. And yes, integrating both sides will always make sense when the function is Riemann integrable.

2) No, we cannot integrate without respect to variable. This would not make sense. You integrated both sides because you already had it in the form involving a differential. You integrated in order to remove the differential.

3) When given $y'' + 5xy = 2x^2$ assuming $y''$ is the second derivative with respect to $x$, you would have to tackle solving this using techniques from ODE, it is not as simple as integrating twice!

4) When you have $xy + y^x = 2x + 3$, you are treating $x$ and $y$ as variables here. You are implying that there may exist solutions, i.e. $x$ and $y$ that satisfy this equation. Clearly, it is not true for every $x$ and $y$, so the right hand side does not equal the left hand side in general. However, there may be certain $x$ and $y$ that make the equality hold true.

  • $\begingroup$ I'm not sure I understand why method 2 would be more appropriate. There's a step in method 2 where I integrate over different variables. $\endgroup$ – Zduff May 23 '17 at 18:20
  • $\begingroup$ Yes, you are integrating both sides. The differential is already there on both sides at that point. The differential tells you what you're integrating with respect to. You are simply putting the integral there making it more natural. Does that make sense? $\endgroup$ – Dragonite May 23 '17 at 18:24
  • $\begingroup$ I think my confusion lies in differentials. I'm not sure what multiplying by a differential means. $\endgroup$ – Zduff May 23 '17 at 18:27
  • $\begingroup$ It is not exactly multiplying by a differential. It is okay to think about it that way. But here is what is going on behind the scenes. $\endgroup$ – Dragonite May 23 '17 at 18:33
  • $\begingroup$ @Zduff FWIW, the mathematicians who invented the notation couldn't tell for centuries what "dx" really meant—though now we have an answer. $\endgroup$ – Blaisorblade May 23 '17 at 22:47

I will disagree with some of the earlier answers a little.

The way we usually do mathematics equality is really an overloaded operator that depends on where we are working. This might not be trivially obvious right away but consider the $2\in \mathbb{N}$, the $2\in \mathbb{Z}$, the $2\in \mathbb{Q}$, the $2 \in \mathbb{R}$ and the $2\in \mathbb{C}$ as an example. Each of these will under standard constructions be completely different objects. Yet we usually prefer to think of them as identical. I.e. If I write $2=(+2)=(2/1)=(\sqrt{2})^2=2+0i$, I would guess everyone would agree these are equal. Even though there is a decent chance the first $2=\{\{\},\{\{\}\}\}$ the second $2=\{1,\{1,2\}\}$ the third $2=\{(a,b);\;a=2b\wedge a,b\in\mathbb{Z}\}$ the fourth $2=\{x;x<2\wedge\; x\in \mathbb{Q}\}$ and the fifth $2=\{2,\{2,0\}\}$, where the symbols $2$ on the right hand sides of the equations should be understood as being the ones defined one earlier.

To answer your questions in order.

1) If two things are equal and you apply the same function to both of them then the result of the application of the function is still equal. In other words if $a=b$ then $f(a)=f(b)$ whenever $f(a)$ makes sense. Now indefinite integration is not a function (that's why you get one of those nasty $c$'s whenever you integrate. Thus performing indefinite integration does not (necessarily) preserve equality. But it does preserve equality up to an additive constant. If $a=b$ then $\int a\;dt=\int b\;dt+C$ for some constant $C$. If all of this is done over the real numbers that constant is a real number. That's why if you write it as you did the last equation holds. The first two don't hold unless you think of the integrals as representing sets of functions. Here I'm talking about method 1. Method 2 is either a lot of abuse of notation or differential one forms. If the former it's the same as method 1, if the latter I don't want to get into that.

2) You can't really integrate without integrating in respect to a variable. (Well you sort of can since you can integrate with respect to a measure but that's more complex and you still need to have some understanding of what the variables are.)

3) Yes up to the caveats given above. You might get things that won't help much but you can integrate.

4) It is not obvious in general from just that equation what you mean. Given no other information than $xy+y^x=2x+3$ there is little we can do with it. It's sorely underdefined, as are most sentences taken out of context. In all we do we try to fill in context. If I say "Trump beats Hillary," you are certainly not given enough context to know what I'm talking about, maybe I know a guy who beats wife and his name if Trump and hers Hillary, or there is a game in which hillary is a card value and trumps beat it. But you will most likely assume that I prefer the current US president to his rival in last years race, since it's the most likely context (I don't necessarily).

The same is with $xy+y^x=2x+3$ you might be asserting this is an identity in which case it's just a false statement. But it's more likely that $x$ and $y$ are variables and you are claiming they have this relationship. It is not obvious what values $x$ and $y$ may take though (is this an equation over the integers? rationals? reals? $\mathbb{Z}_5$?)

To answer your last question I learned what $=$ meant over many years starting in elementary school and mostly having grasped it by the end of my undergrad. Where I went to school I feel there weren't too many obstacles placed in my way to get to the correct understanding. From what I've seen of the US school system it seems to go out of it's way to make sure the students never get the idea.

  • $\begingroup$ In standard first-order logics and set theory (and most other foundations, except ones using setoids), equality is not overloaded (and that's not really helpful here). There's a built-in equality, such that equals can be substituted. So the natural 2 and the integer 2 are not equal, because you can tell them apart. With the set-theoretic encodings you show, "P(x) = {} ∈ x" distinguishes natural 2 and integer 2. $\endgroup$ – Blaisorblade May 23 '17 at 22:57
  • $\begingroup$ There's rather an embedding of naturals into integers, integers into rationals and so on. So you can think that the syntax for numbers is overloaded. For most math in school till undergrad, you can read 2 as the real (or complex) 2 and be fine. Context can dictate otherwise. $\endgroup$ – Blaisorblade May 23 '17 at 22:59

1) An equality always means that the "thing" on the left hand side is the same "thing" as the "thing" on the right hand side. There is no difference in any respect whatsoever between the two "things". If there is some difference, then they are not equal and the equation is incorrect. Therefore, integrating one side of a correct equation must be equivalent to integrating the other side.

2) No. Integration has no meaning at all apart from integrating over some variable.

3) You can write down an integral equation, sure. It won't tell you much without a further solution because of that term that mixes $x$ and $y$. It will just look like $\int 5xy dx$. You simply cannot say that it equals $5yx^2/2$, just in case that's what you were thinking.

4) You are saying that the left and right sides are the same, and then you have to admit the logical implication that this can only be true for the correct combination(s) of $x$ and $y$. If it turns out that there are no such combinations, then you have written an incorrect equation.


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.