# I Question regarding integration of an equation

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1. Jul 4, 2017

I am beginning to learn about differential equations and I saw in an explanatory video a solution to this separable differential equation:
$\frac {dy} {dx} = \frac {-x} {y{e^{x^2}}}$
from there through simple steps the equation changed to $y⋅dy=-xe^{-x^2}⋅dx$.
Then the video did an operation that perplexed me, he said he "integrates both sides of the equation", he then added ∫ to the left side of both sides of the equation to create $∫ydy=∫-xe^{x^2}dx$ and proceeded to integrate both sides of the equation with respect to their respective variables.
I am new to the subject so I may need some elementary explanations on the nature of differentials but few things confuse me here:

First, the dy and dx in this equation are treated as algebraic expressions, so how come adding the integral swirly sign to the left of the expression suddenly turns them into part of the integral notation? If this was a definite integral I'd understand since a definite integral is a sum of these expressions times the value of the functions, meaning that they are still treated as algebraic expressions ($\sum_{n=0}^{→\infty} F(x)dx$). However in this indefinite integral why does this still work? In other words what does he actually do by adding ∫ to the side of the equation?

Second, assuming he can take the integral of both sides of the equation like that, why is he allowed to take the integrals with respect to different variables on the same equation? The left side of the equation is integrated with respect to y and the right side with respect to x, so how do we know that the anti-derivatives are equal?

Sorry for the long question, I would very much appreciate if someone could clear things up for me.

Last edited: Jul 4, 2017
2. Jul 4, 2017

### Staff: Mentor

If it makes you feel better, you can write your equation as $y⋅\frac {dy}{dx} dx = -xe^{-x^2}⋅dx$.
Then integrate both sides with respect to x: $\int y⋅\frac {dy}{dx} dx = \int -xe^{-x^2}⋅dx$.
On the left side, notice that $\frac d{dx} y^2 = 2y \cdot \frac {dy}{dx}$, using the chain rule. This should help you find an antiderivative for the integral on the left.

3. Jul 5, 2017

I see, that explains my second question. However my first question remains unanswered, how come adding the integral swirly sign to the left of the expression turned the dx/dy from an algebraic expression to part of the notation of the integral? Is the indefinite integral also some sort of sum including the multiplication of the function by the differential of its variable?

The way I see it, if $f$ is a derivative of some function $F=y$, then $f=\frac {dy} {dx}$. Multiplying by dx results in $f⋅dx=dy$. Does this mean $dy=y$?

4. Jul 5, 2017

### BvU

No. The derivative times $dx$ is the change in $F$ as 'predicted' by a linearization . $f\, dx = dy$ is not $y$ but the change in $F$.
Make a sketch.

5. Jul 5, 2017

Yes, I thought as much. Meaning both sides of the presented equation are changes in the anti-derivatives (since they are the derivative times dx or dy). If so how does he go from there into finding the anti-derivative? Why does adding the integral swirly sign change the expression of $f⋅dx$, which is merely two algebraic expression multiplied, into the expression $∫fdx$, where only $f$ is an algebraic expression and $dx$ is part of the integral notation? there must be a connection between the two for this change to be possible I just cannot see what it is.
I suppose what i'm really asking here is what is the meaning of $dx$ in the indefinite integral?

6. Jul 5, 2017

### BvU

$dx$ is an infinitesimal. Think of it as a $\Delta x$ that can be made as small as desired (while still $\ne 0$) .

The $\int .... dx$ is the limit of a $\sum ... \Delta x$ with $\Delta x \downarrow 0$. Google Riemann sum and fundamental theorem of calculus

PS this is a physicist's view, don't shock mathematicians with it

No, the $dx$ makes it special.

7. Jul 5, 2017

So, since the expressions on the left and right side of the equation are equal, the sums of the expressions are equal assuming you sum the same way on both sides of the equation (if $ax=by$ then $∑_1ax$ =$∑_2by$ as long as $∑_1$ is the same sum as $∑_2$). And since the fundamental theorem of calculus tells us the indefinite integral at a certain point is just a Riemann sum like that, integrating both sides of the equation w.r.t their variables will yield equal functions (if $f(y)dy=g(x)dx$ then $\sum_{n=0}^\infty f(y)dy=\sum_{n=0}^\infty g(x)dx$ where $\sum_{n=0}^\infty g(t)dt$ denotes the area under the curve of g(t) between the values 0 and t, aka the anti-derivative of g(x) (or f(y) if its the sum on the left side of the equation) at t.
Did I get that right?

8. Jul 5, 2017

### BvU

Getting close.
Some details: (read the riemann story !)

Not $$\sum_{n=0}^\infty f(y)dy$$ but $$\lim_{N\rightarrow\infty} \sum_{n=0}^{N-1} f(y_n) \Delta y$$ with $\ \ \displaystyle { \Delta y = {b-a\over N} }\ \$ and $\ \ { y_n = a + (n+{1\over 2})\Delta y} \ \$ (middle sum as example)

9. Jul 5, 2017

The sum only approaches infinity then, and I assume a and b are the border values. I think I get it now. Thanks for the help.

10. Jul 5, 2017

### Staff: Mentor

Sums are involved only with definite integrals.
You started with $\frac {dy} {dx} = \frac {-x} {y{e^{x^2}}}$. By separating variables (y and dy on one side, and x and dx on the other side), you obtained $y⋅dy = -xe^{-x^2}⋅dx$. The next step is to recognize that if two functions are equal then their antiderivatives will be equal (up to an arbitrary constant).

In an indefinite integral the role of dx is to indicate which is the variable of integration. So, for example $\int x^2 dx = \frac 1 3 x^3 + C$, but $\int y^2 dx$ can't be determined without knowing how y is related to x.

In my previous post I suggested that you could look at this problem as $\int y⋅\frac {dy}{dx} dx = \int -xe^{-x^2}⋅dx$. Just find an antiderivative of each side. There is no summing involved.

11. Jul 5, 2017

But If the role of dx is merely to indicate something, meaning it's simply a notation, how can it be an algebraic expression in the equation. $-xe^-{x^2}⋅dx$ means you multiply the expression $-xe^-{x^2}$ by the expression $dx$, it is a part of the equation. You can't just turn it into a simple notation without justification.
And from my understanding of BvU the justification is that it is a sum, since the fundamental theorem of calculus tells us the indefinite integral is the area function under the curve between a point x and a boundary a in a continuous interval [a,b] $(∫f(x)\, dx=\int_a^x f(x) \, dx=lim_{N\rightarrow\infty} \sum_{n=0}^{N-1} f(x_n) dx)$
If that's not the case or there is another explanation I would love to hear it but otherwise I cannot accept the fact that the algebraic expression dx suddenly turns into a part of the integral notation who's purpose is making a statement without justification.

12. Jul 5, 2017

### Staff: Mentor

In an indefinite integral such as $\int f(x) dx$, the dx factor is simply notation.
What you're talking about is not an indefinite integral -- $\int_a^x f(t) dt$. This is a definite integral (that happens to represent a function of x, the variable that appears as one of the integration limits.

In this thread, we're talking about indefinite integrals, $\int f(r) dr$, or $\int f(w)dw$, whatever. There is no summation going on.
As I already said, the purpose is to indicate the variable of integration.

13. Jul 5, 2017

But the definite integral I described and the value of the indefinite integral at that point are the same function up to an arbitrary constant, as dictated by the fundamental theorem of calculus https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus. Have I misunderstood what the fundamental theorem claims?
Also, you said yet again that the factor is merely a notation, not an algebraic expression. But in the equation $dx$ and $dy$ are clearly treated as algebraic expressions, as they are multiplied and added to other variables in the equation. So why does adding the integral sign to the left of the expressions suddenly deprive them of their status as algebraic expressions and turn them into notations? If they were just notations to begin with why could we multiply, divide, add, and subtract them with other variables as if they were algebraic expressions?

14. Jul 5, 2017

### Staff: Mentor

No, but you're making things more complicated than they need to be.
We're getting way off track here, with discussions about definite integrals and the FTC.
Do you agree that these two indefinite integrals are somehow different: $\int x^2 dx$ and $\int y^2 dx$?
In an indefinite integral, the $\int$ symbol indicates that we are looking for an antiderivative of the integrand (which does not include the differential). Further, the differential dx in both integrals doesn't play a role in the computation, but is nevertheless an important part.

In the study of differential equations, one of the first techniques presented is separation of variables. For example, the differential equation y' = x/y can be solved using this technique, as follows:
$y' = \frac x y$
$\Rightarrow \frac{dy}{dx} = \frac x y$
$\Rightarrow y \cdot dy = x \cdot dx$
The equation is not separated.
If the two expressions making up this equation are equal, then their antiderivatives must be equal, up to a constant.
On the left, the "y antiderivative" of y is $\frac {y^2} 2$. On the right, the "x antiderivative" of x is $\frac {x^2} 2$.
So we have $\frac {y^2} 2 = \frac {x^2} 2 + C$, or equivalently, $y^2 = x^2 + C'$, where C' = 2C.

I can flesh out what I wrote in words above, using indefinite integrals like this:
$\int y \cdot dy = \int x \cdot dx \Rightarrow \frac{y^2} 2 = \frac {x^2} 2 + C$

As a check, I can differentiate both sides of the final equation, with respect to x, getting $\frac {2y} 2 \cdot \frac {dy}{dx} = \frac {2x} 2 \Rightarrow \frac {dy}{dx} = \frac x y$

If the terms "y antiderivative" and "x antiderivate" seem too loose for you, we can make this change $\int y \cdot dy = \int x \cdot dx \Rightarrow \int y \cdot \frac {dy} {dx} \cdot dx = \int x \cdot dx$. Again, all I need to do is to find antiderivatives on both sides.

Writing the integrals as definite integrals makes things needlessly complicated, IMO, and gets away from the idea that all you really want are a couple of antiderivatives.

15. Jul 5, 2017

This is where I get confused. Obviously if 2 expressions are equal their anti-derivatives are equal up to a constant. However the derivatives you presented are of $y$ ($\frac {y^2} 2+c$) and $x$ ($\frac {x^2} 2+c$), and this would have been fine if the equation you presented was $y=x$, but it isn't, you presented the equation $y⋅dy=x⋅dx$, so taking the integral of both sides of the equation would be $∫y⋅dy\, dy=∫x⋅dx\, dx$, wouldn't it? These are the anti-derivatives of the sides of the equation you presented.

Writing the integrals as definite integrals was of course needlessly complicated, but it in order to understand the reason for the fact the $dx$ and $dy$ turned from algebraic expressions to notations, since if we treat the integrals as definite integrals they always remain algebraic expressions. It was merely to make me understand the reason behind a certain operation. But if we ignore this statement then how come you can suddenly turn the expressions of $dx$ and $dy$ from algebraic expressions that can be devided, multiplied, added, subtracted and so on with other expressions into notations which cannot do all those things and just exist in order to make a statement?

16. Jul 5, 2017

### Staff: Mentor

You missed something I wrote:
The (indefinite) integrals in the equation on the right are both integrated with respect to x.
In this integral, $\int y \cdot \frac {dy} {dx} \cdot dx$, I have multiplied by $\frac{dx}{dx}$, or 1.
An (x) antiderivative of $y \frac{dy}{dx}$ is $\frac 1 2 y^2$, working the chain rule backwards.

17. Jul 5, 2017

It seems there is a bit of missunderstanding regarding what I am actually asking, so allow me to clear up my question to the best of my abilities:
I understand what you did, my problem was not with integrating the sides of the equation w.r.t. different variables, that was addressed in your first reply. I understand how you can differentiate different sides of an equation with respect to different variables.
My problem is how you suddenly change the $dx$ and $dy$ from algebraic expressions into notation. At first you multiply the equation by $dy$ and $dx$, implying that they are algebraic expressions. Then you add the integral sign the completely ignore them as algebraic expressions, instead suddenly treating them as notations who's single purpose is to show with respect to what variable the integral is being made.
If all the $dx$ and $dy$ are are notations with the purpose of claiming with respect to what variable a function is being integrated or differentiated, then what is the meaning in the equation $y⋅dy=x⋅dx$? The equation here does not contain a derivative of integral, then what is the meaning of $dx$ and $dy$?

If you integrate the equation $y=x$, the left side with respect to y and the right side with respect to x, you create the equation $∫y\, dy=∫x\, dx$, If you integrate the equation $2y=3x^2$, the left side with respect to y and the right side with respect to x, you create the equation $∫2y\, dy=∫3x^2\, dx$. Similarly, if you integrate the equation $a=b$ where $a$ and $b$ are any expressions, the left side with respect to y and the right side with respect to x, you create the equation $∫a\, dy=∫b\, dx$. In this case $a=y⋅dy$ and $b=x⋅dx$, thus, integrating the equation, the left side with respect to y and the right side with respect to x should produce the equation $∫y⋅dy\, dy=∫x⋅dx\, dx$.

However, instead of doing that you decided that $dy$ and $dx$ are no longer part of the algebraic expression itself, but instead simple notations who share the same meaning as the $dy$ and $dx$ in any integral, despite the fact that you treated them as algebraic expressions representing something as you multiplied the equation by $dx$ and $dy$ in the previous step.

So my question is simple, if the $dx$ and $dy$ are just notations, as you said: "In an indefinite integral such as $∫f(x)\, dx$ the $dx$ factor is simply notation", then how come you can multiply variables like x and y with these notations? How come you can multiply an equation by $dx$ and $dy$ the same way you multiply an equation by $x$ or by 2 if they are just notations unlike $x$ and 2 which are algebraic expressions like constants and variables?

If you can do that, what is the algebraic meaning behind doing so? When you multiply an equation by 2 it means both sides of the equation are now twice what they were before the operation, when you multiply an equation by $x$ it means both sides of the equation are now multiplied by the variable $x$. So when you multiply an equation by $dx$ (or $dy$) that means both sides of the equation are multiplied by what? A notation of an integral and derivative? how does it affect them?

Hope that clears things up.

18. Jul 5, 2017

### Staff: Mentor

It is simply a rearrangement of the equation $\frac{dy}{dx} = \frac x y$.
Without knowing how a is related to y and how b is related by x, the antiderivatives of the above equation can't be found.
First, it makes no sense to have a = y dy, and replace a with this expression in $\int a dy$ (getting $\int y~dy~dy$). This is meaningless.
They are notation, but that isn't to say that they are redundant and don't have meaning.
Some textbooks write indefinite integrals as $\int f$, with the implied meaning that f is a function of some variable, and the integration is done with respect to that variable. It would be meaningless to write $\int x^2y$ without indicating what the variable of integration is.
Your question is not clear. You can multiply both sides of an equation by dx or you can multiply both sides by dy, just the same as you would expect. Are you asking whether we can multiply one side by dx and the other side by dy? If so, that's not valid.
Both sides are multiplied by dx (or dy, if that's what you multiply both sides by).
Example:
$\frac {dy}{dx} = x$ - a differential equation
Multiply both sides by dx to get
$dy = x~dx$
Integrate:
$\int dy = \int x~dx$
On the left, the integrand is 1. What is the function of y whose derivative is 1? I.e., $\frac d {dy}(?) = 1$
On the right, the integrand is x. What is the function of x whose derivative is x? I.e., $\frac d {dx}{?} = x$

Clearly, we get y = (1/2) x2 + C.
As a check, dy/dx = x, so the solution (actually an entire family of solutions) satisfies our original differential equation.

Solving a differential equation means undoing the differentiation process by antidifferentiating to find the original function that was involved in the diff. equation.

Last edited: Jul 5, 2017
19. Jul 5, 2017

My question remains unanswered, I am not asking what you are doing or how you are doing it, I understand that perfectly, I am asking for the reason that those things can be done. Everything in mathematics has a reason, I want to know the reasons behind these actions. These expressions represent functions, all operations that are done must agree with those functions. The $\frac {dy} {dx}$ are not just notations that say "this is the derivative of y with respect to $x$, the $dy$ is $\Delta x$ as it approaches 0, the $dy$ is $\Delta y$ as it approaches 0 in correspondence with $dx$, and the whole expression is the limit of a slope. And this value of this limit is the value of the derivative at that point, it's not just notation, it is a proper algebraic expression whose value is represented by the derivative function.
That is what $dx$ is, the limit of $\Delta x$ as it approaches 0, it is an expression, $\frac {dy} {dx}$ is also an expression, and when you set x as a certain value it receives a value too.

Thus the equation $dy=x⋅dx$ is also comprised of algebraic expressions, these are $dy$, $x$ and $dx$. So when you say you add the integral sign to the left of the expression without doing anything to $dx$ this means $dx$ remains the same thing, an expression for the limit of $\Delta x$ as it approaches 0.

I wished to know then, can adding the integral sign to the left of the equation suddenly makes it a part of the notation of the indefinite integral. In short I asked what is the meaning of $dx$ in the indefinite integral, why did the mathematicians of old that came up with these notations decided to make it as so? My question was answered after delving into the fundamental theorem of calculus seeing that the $dx$ is a part of a sum, the sum described in the FTC, and thus it retains its one and only meaning as the limit of $\Delta x$ as it approaches 0. I just wanted to know the "why" and my question was answered, so I appreciate the help of both mark44 and BvU.

20. Jul 5, 2017

### Staff: Mentor

You have a typo: "the $dy$ is $\Delta x$ as it approaches 0"
Obviously you meant "dx is $\Delta x$ as it approaches 0". This doesn't make a lot of sense, though, since $\lim_{\Delta x \to 0} \Delta x$ is just zero.
Usually dx is defined as "an infinitesimally small change in x" and dy is defined in terms of f'(x) and dx. I.e., dy = f'(x) dx.

Possibly you're struggling with something that is taught at the beginning of calculus, that $\frac {dy}{dx}$ (the derivative of y with respect to x in Leibniz notation) should not be thought of as a fraction. However, following calculus, in studying differential equations, you are taught that there's no problems with treating the ratio exactly as you would an ordinary fraction.

That's not what I said ("add the integral sign"). What I said was that we can antidifferentiate both sides of the equation dy = x dx. The notation for this includes the integral sign. If you have two expressions that are equal, you can differentiate both sides to get a new equation, or you can antidifferentiate both sides to get a new equation.

Because it is suggestive of $\Delta x$ in a Riemann sum. For indefinite integrals, it serves more as notation that indicates what kind of antiderivate we want. In an indefinite integral, you aren't adding up the areas of a bunch of rectantles or other figures.
Again, this isn't how dx is defined. Think about this: what is $\lim_{h \to 0} h$? It's difficult to justify any answer other than zero.