d/dt: operator or variable?

[Allan_Ecker]

I'm trying to re-learn DiffEq from scratch to fill in holes in my knowledge, and I seem to be running into this identity that to me makes no sense. I ran into it rather early and so kind of drew a box around it and moved on, but I can't help but think it's really important to fundamental understanding:

The Sixth Edition Boyce&Diprima Elementary Differential Equations and Boundary Value Problems text solves the following first-order equation:

dy/dt + y/2 = 3/2

in a way that really confuses me. To better illustrate exactly what's confusing me, I'll use a simplified equation which has the same point of contention:

dy/dt = y

Which is equivalent to:

dy/dt * 1/y = 1

Which my textbook jumps to:

d/dt ln(y) = 1

There's some text, indicating this was done via differentiation:

dy/dt * 1/y = 1
dy/dt * (d/dy ln(y)) = 1
d/dt ln(y) = 1

Above is where I get confused. I've lived my life thus far with a blurry notion that dy and dt were standins for delta-y and delta-t, where delta is really "the limit as delta goes to zero" and if dy and dt are variables, then in the above steps dy cross-cancels with itself to yield the final answer of d/dt. But if dy is a variable, what is d/dt? Is "d" a variable? Or is d/dt an operator? If d/dt is an operator, isn't d/dy an operator too? If they're both operators and not variables, then how is this cross-multiplying step possible? Is that an identity I don't know about? Such an identity would be:

d/dt(y) * d/dy(f(y)) = d/dt(f(y))

Which then sort of slams me face-first into the fact that all the calculus identities I know are monovariate identities which don't really apply here. Help?

 

[tkhunny]

The notation "d/dt" is a little ambiguous. It should be understood to be an OPERATOR ONLY in this form. It CANNOT stand alone in this form. A function argument must follow.

\(\displaystyle \frac{d}{dt}f(t)\;=\;\frac{df}{dt}\)

If we are given the definition of f(t) = 3t - 5, we then have:

\(\displaystyle \frac{d}{dt}(3t-5)\;=\;\frac{d(3t-5)}{dt}\;=\;3\)

It is okay to have dual meanings. \(\displaystyle \frac{dy}{dx}\) certainly can mean the derivative of y with respect to x. Standing there by tiself, what else would it mean? But in an equation, it could change nature. \(\displaystyle \frac{dy}{dx}\;=\;\frac{x^{2}}{y}\;\rightarrow\;y \cdot dy\;=\;x^{2} \cdot dx\). Again, this should be a little uncomfortable unless we make the extention that both x and y are functions of some other variable, often 't'.

Whether the notation is a derivative, representing only the definition of a derivative, or the notation means a quotient of two differentials, it makes a difference only to your application. It is a far cry from the earlier "infinitesimals". But that's another story.

Well, that's just the insta-version. Probably nothing you haven't seen already.

 

[Allan_Ecker]

Okay, that's a usable definition for now, I hope. The next chapter deals with differential equations which are "separable" which I think has something to do with the split of dy/dx into dy and dx, as a ratio. I think I'm facing a little bit of a fundamental problem in that the real definitions of a lot of these things are buried in mathematics considerably scarier than the demure differential equations I'm dealing with, kind of like how I spent a few years living in secret terror that someone would ask me to actually *show* that the derivative of x^2 was 2x using limits.

(Actually it'd probably take me an afternoon to get it right even now, barring trips to Wikipedia and/or Wolfram, but still.)

This all has me worrying that the curly d's associated with PDEs have some meaning beyond flags that they exist in a multivariate framework...

Well as Dr. Allstot tends to say, We'll burn that bridge when we come to it.

 

[skatty14]

Thank you, I have been secretly wondering what the devil they were trying to say by using that notation. I find the apostrophe notation less confusing, but I suppose you can't split it up and therefore can't do separable equations. I also don't really understand limits. I know that if you have a graph that has say a horizontal asymptote on y = 1, then the limit of x > infinity = 1, but when they delve into the calculus of it all, then I am totally lost! Wish they would be less technical at first and just tell us wtf they are trying to say! Even though I am a math major I too am often confused, you are not alone!!! Still confused, but that helped a bit, thanks.

 

[Someone2841]

It may be helpful to note that \(\displaystyle \frac{df(x)}{dx}\) is shorthand for \(\displaystyle \lim_{\Delta \to 0}\frac{f(x+\Delta)-f(x)}{\Delta}\)

Therefore, one can (if one is careful) treat \(\displaystyle df(x)\) to mean \(\displaystyle \lim_{\Delta \to 0}(f(x+\Delta)-f(x))\).


For example, consider the equation \(\displaystyle y = f(x)\) and its inverse \(\displaystyle x = g(y)\). This means:

\(\displaystyle f'(x) := df(x)/dx = \lim_{\Delta \to 0}\frac{f(x+\Delta)-f(x)}{\Delta}\)

\(\displaystyle g'(y) := dg(y)/dy = \lim_{\Delta \to 0}\frac{g(y+\Delta)-g(y)}{\Delta}\)


From these, we can derive that \(\displaystyle 1/f'(x) = g'(y)\). That is, the reciprocal of a function's derivative is equal to the derivative of its inverse function:

\(\displaystyle f'(x) = \lim_{\Delta \to 0}\frac{f(x+\Delta)-f(x)}{g(y+\Delta)-f(y)} \to 1/f'(x) = \lim_{\Delta \to 0}\frac{g(y+\Delta)-f(y)}{f(x+\Delta)-f(x)} = g'(x)\)


Using short hand:

\(\displaystyle f'(x) = \frac{dy}{dx} = \frac{df(x)}{dg(y)} \to 1/f'(x) = \dfrac{1}{\frac{df(x)}{dg(y)}} = \frac{dg(y)}{df(x)} = \frac{dx}{dy} = g'(y)\)



Another example is the technique you use to solve first order separable equations. This is all to say, while \(\displaystyle \frac{d}{dx}\) is often used as an operator, \(\displaystyle dx \) and \(\displaystyle dy\) do have mathematical meaning apart from each other (i.e., apart from in the form \(\displaystyle \frac{dy}{dx}\))

 

[kevin786]

Opertor or variable

I tried to solve this questions but unfortunately i couldn't make it comparable here.I will consult my friend for this one and i will definitely answer this one very shortly.
Will keep eye on this one,will surely help you out.