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Question about dy/dt

  1. Feb 4, 2005 #1
    First of all I realize that this question has been asked before and answered before and so I appolgize if I am making any of you feel like you're talking to yourselves. But I am curious and the people here are the smartest people I know and so I want to hear from you guys and gals, an answer for this age old question.

    When we have an expression like [tex]\frac{dy}{dt}=y+5[/tex] and we are asked to find a solution to this DE we

    treat the numerator and denominator as if they were normal functions. How am I to think of this? I mean at a certain values of y and t [tex]\frac{dy}{dt}[/tex] is infact a number.

    Oh, bother..... :mad: I don't know how to ask the question correctly. What is this dy/dt thing? A change in y over a change in t or a quotent of funcitons or what?

    Any help is appreciated very much. I have no problem working the problems from the text but eventually I will need to really understand what is going on so perhaps this is a good place to start.

  2. jcsd
  3. Feb 4, 2005 #2


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    IT IS A FUNCTION.It is unknown,just like the functions "y" itself...

  4. Feb 4, 2005 #3


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    If y is some function of t, given by :

    [tex]y = f(t) [/tex]

    Then, the derivative of y (w. r. to t) is defined by

    [tex]\frac{dy}{dt} \equiv f'(t) = \lim_{\Delta t \rightarrow 0} \frac {f(t + \Delta t) - f(t)}{\Delta t} [/tex]
  5. Feb 4, 2005 #4
    Ok so
    This means the solution is just y=t+c. But to get there we multiplied both sides of the equation by dt. So dy=dt. So what is dy? A change y in for a change in t? How would I graph this function then? What does its graph look like?

    I don't know, I guess I always thought there was more to it then that. Since a derivate of a function is found using the limit process then I figured it would be something more like

    [tex]\frac{dy}{dt}=lim_{h \rightarrow 0} \frac{f(t+h)-f(t)}{h}=1[/tex]

    So then dy/dt is not really a quotient of functions that I can see. so how is it that we treat it like one? I mean in terms of limits what does dy=dt look like?
    Last edited: Feb 4, 2005
  6. Feb 4, 2005 #5
    By the way the function I am using for an example is of course y(t)=t+c. Just in case someone was wondering why I am saying dy/dt=1 and it was not clear from what I was saying.

    In any case I have to goto class so this will have to wait.

    I'll be back...

  7. Feb 4, 2005 #6


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    Plot it and see.

    I think you've just written how [itex]\frac{dy}{dt}=1 [/itex] looks like in terms of limits...
  8. Feb 4, 2005 #7

    matt grime

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    dy isn't anything. We use it as a symbol to tell us what to do, and it behaves in some isolated cases as if we could separate dy/dtlike a fraction BUT IT ISN'T A FRACTION.

    If f(y)dy/dt = g(t), then it follows that int f(y)dy = int g(t)dt

    dy and dt are not themselves things that you need to worry about, they are just symbols.

    Later on you will learn about differentials, but you do not need those right now.
  9. Feb 4, 2005 #8


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    The point is that dy/dx is NOT a fraction but it is a limit of a fraction so can be "treated" like a fraction: to prove a "fraction" theorem about dy/dx, go back before the limit, use the fact that the "difference quotient" is a fraction, and then take the limit again.

    In order to use that we define the "differential" dx as purely notational and then define the corresponding dy by dy= f '(x)dx (the only time it makes sense to have a "dx" without a corresponding "dy" or vice-versa is when you are integrating).

    When you get into functions of several variables, differentials get a whole lot more interesting!
  10. Feb 4, 2005 #9


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    Like the others said. dy/dt is not a fraction, and in that sense, writing dy/dt=1 as dy=dt and then integrate is a fraudulent way to get the solution of the DE.
    For starters, it may be best to think about it as follows. Although it's strictly not allowed. Let's try it anyway. That's the nice thing about differential equations. It doesn't matter HOW you get your answer. You can check whether it is correct afterwards anyway by putting it in the DE.

    That being said, the reason why it works because of the substitution rule (which in turn is a result of the chain rule of differentiation).
    Note that we can only seperate the dy and dt and integrate when we can put the stuff depending on y on one side and the stuff depending on t on the other side. The equation is then seperable.
    It is then always of the form: f(y)dy/dt=g(t), for some functions f of y and g of t. Now if we integrate both sides wrt t:

    [tex]\int f(y)\frac{dy}{dt}dt=\int g(t) dt[/tex]

    by the substitution rule, this equals:
    [tex]\int f(y)dy = \int g(t)dt[/tex]

    which is exactly the same result you get by treating dy/dt as a fraction.
    It is therefore justified to treat them like fractions in this case.
    Last edited: Feb 4, 2005
  11. Feb 4, 2005 #10
    what you said was right, it is the change in y with respect to the change in t.
    we limit t to become a very small number, and see what difference is it compared with y in a given function. because dy = dt, then you will get a graph where a small change in y (within the y axis) will give an EQUIVALENT small change in t (in the t axis)
    i believe it is the same as saying y = t, which is just a unit of y compared with a unit of t. well, it can be proven using integration as well.
  12. Feb 4, 2005 #11
    Everyone, thankyou for your replies....That being said Galileo, your reply really made things clear for me. I can relate what you said to the chain rule and actually I don't like to use a u sub to solve integrals. I try to see things in terms of the chain rule.


    Good thing we have the chain rule to use. It seems like so much of calculus depends on it.

  13. Feb 4, 2005 #12
    This is the best explanation I could come up with which is composed of ideas from various calc books, and from the explanations of some of the people here at PF. I would appreciate it if someone could look this over and make sure its right.

    Just to add to what has been said here...
    As Galileo and others have mentioned, when we treat [itex] \frac{dy}{dx}[/itex] as if it were a fraction to solve differential equations, we use this simply as a notational device to hide the messy details of the chain rule.

    To give an example:
    suppose [itex] y = f(x) [/itex] ,

    Let [itex]g[/itex] be a function of [itex]x[/itex] and [itex]G[/itex] be an antiderivative of [itex]g[/itex], therefore...

    [tex] \frac{d}{dx}G(x) = g(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex]

    Let [itex]h[/itex] be a function of [itex]y[/itex], and [itex]H[/itex] be an antiderivative of [itex]h[/itex], therefore...

    [tex] \frac{d}{dy}H(y) = h(y) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)[/tex]

    also realize that [itex]H[/itex] and [itex]h[/itex] are both composite functions of [itex]x[/itex] (since [itex]y = f(x)[/itex])
    therefore (2) can be written as

    [tex] \frac{d}{dy}H(f(x)) = h(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)[/tex]

    we can use the chain rule on (3) to find the derivative of [itex]H[/itex] wrt [itex]x[/itex],

    [tex] \frac{d}{dx}H(f(x)) = H'(f(x))f'(x) [/tex]

    but from (2), [itex] H'(f(x)) = h(f(x))[/itex] so this becomes

    [tex] \frac{d}{dx}H(f(x)) = h(f(x))f'(x) \ \ \ \ \ \ \ \ \ (4) [/tex]

    Now here comes an example of how the Liebeniz notatin is useful when we reverse the process and are given a derivative and are asked to find the original function ( A differential equation ).

    suppose were asked to solve:
    [tex] \frac{dy}{dx} = \frac{g(x)}{h(y)} [/tex]

    we usually use the method of seperation of variables withouth hesitation to rewrite this as

    [tex] h(y)dy = g(x)dx [/tex]

    and solve

    [tex] \int h(y)dy = \int g(x)dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)[/tex]

    since [itex]H(y)[/itex] is an antiderivative of [itex]h(y)[/itex]
    and [itex]G(x)[/itex] is an atiderivative of [itex]g(x)[/itex]
    the solution to (5) is

    [tex] H(y) = G(x) + C \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)[/tex]

    now it looks like we were able to treat [itex]dy[/itex] and [itex]dx[/itex] as if they were components of a fraction, but if we look under the covers, its just the chain rule in reverse. We can see this by rewriting the LHS of (5) subsituting [itex]y = f(x) [/itex] and [itex]dy = f'(x)dx. [/itex]

    [tex] \int h(f(x))f'(x)dx = \int g(x)dx \ \ \ \ \ (7) [/tex]

    our goal is still to find the antiderivative of each side

    but looking at (4) we see that [itex]H(f(x))[/itex] is an antidiverative of [itex]h(f(x))f'(x)[/itex] ( this is just the reverse of the chain rule ).

    Thus our solution is

    [tex] H(f(x)) = G(x) + C [/tex]

    again, subsituting y = f(x) we get the same result

    [tex] H(y) = G(x) + C [/tex]

    so you see,,, Libeniz's notational device is an easier method to solve such problems because it lets us hide all the messy details of the chain rule.

    There is also another common use for Liebeniz's differential notation which allows us to approximate changes in a function [itex] \Delta y [/itex] with a corresponding change in the indepent variable(s) [itex] \Delta x [/itex]. This method is known as the linear approximation, and is given by

    [tex] dy = f'(x)dx [/tex]

    where it is defined that [itex] dx = \Delta x [/itex]

    here [itex]dy[/itex] and [itex]dx[/itex] are assigned values, and we use dy to approximate the value of [itex] \Delta y [/itex], ( where the approximation gets better, the smaller the [itex] \Delta x [/itex] ).

    These two devices for the Liebeniz notation are not to be confused with each other, they are used for seperate purposes.

    Let me know if this is any help. I tried to make it as explicit as possible, but I'm afraid it might have become too lengthy in the process. Also please correct me if I am wrong somewhere or if you think there is more that should be added.

    Last edited: Feb 4, 2005
  14. Feb 4, 2005 #13
    I am not sure number 3 is really correct, or maybe I am not seeing it right...

    But overall everything looks really good and you managed to make the undercover work very explicit.

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