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I'm starting to learn about differential equations, and I'm having trouble mentally adjusting to working with differentials as separate quantities. (I took calculus in high school and college but I don't remember ever learning about differentials or differential equations, so the whole concept is refusing to stick in my brain.)

Mechanically I can do the work and get the right answer. But I don't really understand why it's correct to do the operations that I'm doing. My problem is that I can't quite grasp why it's OK to do arithmetic on differentials just because Leibniz's notation happens to make it "look right". I mostly get (at least, I can accept) why we can take a DE like this:

$dy/dx = -xy$

and multiply through by $dx$, divide by $y$, then anti-differentiate the resulting two parts. I have always thought that $dy/dx$ was just a different form of notation for a function, $f'(x)$ that described the rate of change of y vs. x. But I can accept that it's actually a ratio of two infinitesimal numeric values.

Where my understanding fails is with this form of the same differential equation:

$(d/dx + x)y = 0$

at which point we "distributed" the $y$ to get the subsequent step. But I have no idea what $d/dx$ actually means. In my mind, I have treated it as an operation applied to functions: you "apply" $d/dx$ to a function written in terms of $x$ to get it's derivative. How, then can we add to and multiply by an operation? To me, that second equation looks exactly as if you had written $(! + x)y = 0$ and expanded that to $y! = -xy$, which obviously makes no sense. Why then, does it "work" for $d/dx$? What numeric quantity is $d/dx$ supposed to be?

KutuluMike
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  • Related: http://math.stackexchange.com/questions/20685/in-differential-calculus-why-is-dy-dx-written-as-d-dx-y – Jonas Meyer Sep 29 '13 at 19:09
  • @JonasMeyer thanks, that looks like interesting information; but I got lost by sentence three of the answer :) I assume that stuff is from linear algebra (next on my list)? – KutuluMike Sep 29 '13 at 19:11
  • Technically, it just notation for a specific limit, or, later, a linear function on a function space. – Thomas Andrews Sep 29 '13 at 19:27
  • This has been asked here before, and great answers were given. – Pedro Sep 29 '13 at 19:38
  • @PedroTamaroff I've read lots of questions/answers that talk about $d/dx$ and all of them reinforce my intuition that it's an operator, not a numeric value you can do arithmetic on... – KutuluMike Sep 30 '13 at 01:22
  • @MichaelEdenfield OK...? – Pedro Sep 30 '13 at 01:23
  • @PedroTamaroff so if my question has been asked and answered on here before I cannot find it, so I would appreciate any pointers... – KutuluMike Sep 30 '13 at 01:41
  • @MichaelEdenfield I was thinking about this but now I realize the question is different so I apologize. You can think of it as a linear operator in some sense. I recommend you take a look at Spiegel's "Applied Differential Equations", specially the section 4.6 called "Abbreviated Methods Involving Operators". – Pedro Sep 30 '13 at 02:04
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    To give a suggestive example, consider the differential equation $y''-y'+y=x^3-3x^2+1$. We can write this as $(D^2-D+1)y=x^3-3x^2+1$. But $$\frac{1}{1-D+D^2}=1+D-D^3-D^4+\cdots$$ whence $y=1(x^3-3x^2+1)+D(x^3-3x^2+1)-D^3(x^3-3x^2+1)-D^4(\cdots)$ but the fourth derivative already vanishes so we get $y=x^3-6x-5$. Some details must be not forgetten. For example, $D$ doesn't commute with functions in the sense $D(fg)\neq fDg$ say. – Pedro Sep 30 '13 at 02:08
  • To only touch on one aspect of your question, let me point out that $d/dx+x$ does make a lot of sense when $d/dx$ is thought of as an operator, if you also see $x$ as an operator ('operation,' as you call it): it's multiplication by $x$. You will see this in Linear Algebra, too, where one frequently encounters expressions of the form $(A-\lambda I)x=0$. Here, $A$ is a matrix but $\lambda$ is a scalar ('numeric value'); to add them up, one employs the identity matrix $I$ (same dimension as $A$). In your ODE example, $x$ is implicitly multiplied by the identity operator - that's all. – automaton 3 Sep 30 '13 at 07:53

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