C1 and C2 in a differential equation....

[luckis11]

d^2x/dt=0.01-0.01dx/dt
=>x(t)=-100c1(e^-0.01t)+c2+t
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

 

[Mark44]

d^2x/dt=0.01-0.01dx/dt
=>x(t)=-100c1(e^-0.01t)+c2+t
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

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).

 

[luckis11]

d^2/dt means ddx/dt and the solution was from wolframalfa, thus you disagree with them. I don't 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.

 

[pasmith]

d^2/dt means ddx/dt and the solution was from wolframalfa, thus you disagree with them.

Woolfram alpha interprets the incorrect and meaningless "d^2x/dt" as the correct "d^2x/dt^2".

 

[Mark44]

d^2/dt means ddx/dt

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 $\frac d {dt} ( \frac{dx} {dt})$.

 

[luckis11]

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.
dddx=d^3x
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

 

[weirdoguy]

My mistake d^2/dt meant d^2x/dt at the second post.

It still does not make sense, whether you like it or not.

 

[Mark44]

From your 2nd post (post #3):

I don't 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.

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:

My mistake d^2/dt meant d^2x/dt at the second post.

Still incorrect. If you mean $\frac d {dt}(\frac {dx}{dt})$, that is $\frac{d^2x}{dt^2}$.