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I C1 and C2 in a differential equation...

  1. Sep 29, 2016 #1
    How do we find c1 and c2. Are they numbers or functions?
    d^2x/dt^2 instead of d^2x/dt gives the same solution, which means different c1 and c2
    Last edited: Sep 29, 2016
  2. jcsd
  3. Sep 29, 2016 #2


    Staff: Mentor

    c1 and c2 are constants (numbers). You can't determine them without having initial conditions.

    Some of your notation is screwy. d^2x/dt doesn't mean anything. It should be d^2x/dt^2, or better yet ##\frac{d^2x}{dt^2}##, which you could write more simply as x''(t).
  4. Sep 30, 2016 #3
    d^2/dt means ddx/dt and the solution was from wolframalfa, thus you disagree with them. I dont have time to reassure that the same solution for d^2x/dt=0.01-0.01dx/dt and for d^2x/dt^2=0.01-0.01dx/dt simply means different c1 and c2.

    From some of my calcultions it seems that in this case c1 c2 are numbers (in other cases I guess they are not), so ignore the initial question.
  5. Oct 1, 2016 #4


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    Homework Helper

    Woolfram alpha interprets the incorrect and meaningless "d^2x/dt" as the correct "d^2x/dt^2".
  6. Oct 2, 2016 #5


    Staff: Mentor

    No it doesn't. As pasmith said, d^2/dt is incorrect and meaningless. Also, ddx/dt is equally incorrect and meaningless.

    The notation ##\frac{d^2x}{dt^2}## is shorthand for ##\frac d {dt} ( \frac{dx} {dt})##.
  7. Oct 3, 2016 #6
    My mistake d^2/dt meant d^2x/dt at the second post. You should leave more time to edit our posts.
    d^2x MEANS ddx. That is, the difference between the last and the previous dx.
    When speed is increasing: ddx/dt=d(dx/dt)=du if all the dt are equal, whereas if all the dx are equal ddx=0
  8. Oct 3, 2016 #7
    It still does not make sense, whether you like it or not.
  9. Oct 3, 2016 #8


    Staff: Mentor

    From your 2nd post (post #3):
    Above, your first equation is incorrect, because of the d^2x/dt term. This has been stated several times. The expression d^2x/dt could be correct, assuming you're talking about the differential of dx/dt, instead of the derivative of dx/dt, but from the context of these equations, you're dealing strictly with the first and second derivatives.
    Personally, for equations such as this, modified Newton notation is simpler to write, either as x'' = .01 - .01x', or more explicitly as x''(t) = .01 - .01x'(t), showing t as the independent variable.
    From post #6:
    Still incorrect. If you mean ##\frac d {dt}(\frac {dx}{dt})##, that is ##\frac{d^2x}{dt^2}##.
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