I believe I understand the meaning of the infinitesimals $dx$ and $dy$. I understand that $dx/dy$ is the ratio of an infinitely small change in $x$ to an infinitly small change in $y$. However, I can not imagine what $(\frac{d}{dx} +x)y=0$ is trying to say.

If it says $\frac{d}{dx}y$ I understand that this would be the notation for the derivative of $y$. However, $\frac{d}{dx}$ is not paired with any real number of which it would be sensible to take the derivative, it is all by itself. What is interesting is that the professor takes the statement $(\frac{d}{dx} +x)y$ and multiplies out $y$, making $\frac{dy}{dx}+xy$ which is a statement that I can make sense of. However, I can not understand how this was a legal move.

Here is a timestamped link to the video in which the problem pops up.

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    $\begingroup$ It's a legal move simply by definition: $dy/dx + xy$ is simply what $(d/dx + x)y$ means. (The meaning of $d/dx$ is the operation of differentiation.) $\endgroup$ Dec 27, 2015 at 17:14
  • $\begingroup$ @DavidC.Ullrich So what $\frac{d}{dx} +x$ is telling me to do is to add the operation of differentiation to the real number x? Is the result a number/variable, or does $\frac{d}{dx} +x$ only make sense when paired with another number/variable? $\endgroup$ Dec 27, 2015 at 17:17
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    $\begingroup$ $\frac d{dx}$ is in this context an algebraic symbol like any normal letter. Distribution works like it normal, and it is associative. It commutes with itself (obviously) and with $\frac d{dy}$, but not with $x$ or $y$. When you have finally removed all brackets, you may begin to interpret it as a differentiation. $\endgroup$
    – Arthur
    Dec 27, 2015 at 17:19
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    $\begingroup$ Yes. Well, actually it's adding the operator of differentiation to the operator "multiply by $x$. Whether "$d/dx+x$" makes sense by itself depends on what sort of sense you want to make of it - it's a perfectly respectable operator... $\endgroup$ Dec 27, 2015 at 17:20
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    $\begingroup$ Take a look at this post which is my favorite! In fact, $\frac{d}{dx}[\,]+x \cdot [\,]$ is just an operator. :) $\endgroup$ Dec 27, 2015 at 17:26

2 Answers 2


Another way to write this is $y'(x) + xy(x)= 0$.

What you are supposed to do is very likely to solve the thus given differential equation. That is determine a function $f(x)$ such that $f'(x) + x f(x) = 0$.

Or it just may be the assertion that a (certain) function is $y(x)$ has that property. In a formal sense it is not all that different from, say, $y^2 + y x= 0$, which is also a equation involving $y$ and $x$ and some operations on these functions. The sole difference is that here you have the additional operation "take derivative."


The symbols $d/dx$ and $x$ should both be interpreted as linear operators acting on a vector space that the unknown function $y$ belongs to. The sum of linear operators is well-defined and that is exactly how the $y$ ends up between or outside the brackets. The first operator acts by differentiation and the second one by pointwise multiplication.

  • $\begingroup$ I have not yet taken linear algebra or whatever subject is required to understand vector spaces. However, I do now understand that I was wrongly thinking about the statement as a sum of two numbers, when in reality I should have considered it as the sum of two operators. $\endgroup$ Dec 27, 2015 at 18:04
  • $\begingroup$ On a side note, in what subject is operator theory usually covered? Calc2, differential equations? Or was I supposed to cover it in some earlier subject from high school? $\endgroup$ Dec 27, 2015 at 18:05
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    $\begingroup$ Depends on the approach. I had it in advanced calculus but differential equations, especially partial ones, are a strong motivator. $\endgroup$ Dec 27, 2015 at 18:33

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