Integral of d²x?

Question

Given a differential equation of the form

$ \ddot x = g $,

the solution of $x$ is given by

$x = \int\int g \cdot dt \cdot dt = \int\int g \cdot (dt)^2 $

However, were the first equation originally in Leibniz notation, the latter equation would read

$ x = \int\int d^2x = \int\int g \cdot (dt)^2 $

is this correct, or is this an abuse of the Leibniz notation? If its not, then it would mean that

$ \int\int d^2 x = x + C $

which is the focus of this question. I can only infer afterwards that

$ \int \int d^2 x = \int (\int dx) d(1) = \int (x+C) d(1) = (x+C) \int d(1) = (x+C) $

Once again, is this a misuse of the notation?

Answer 1

Liebniz notation is great, but the second derivative notation is actually generally written wrong. See the paper Extending the Algebraic Manipulability of Differentials.

A better way to look at it is to say that, assuming $g$ is constant:

$$ \frac{d\left(\frac{dx}{dt}\right)}{dt} = g $$

Multiplying both sides by $dt$ yields:

$$ d\left(\frac{dx}{dt}\right) = g\,dt$$

Integrating both sides gives:

$$ \int d\left(\frac{dx}{dt}\right) = \int g\,dt \\ \frac{dx}{dt} = gt + C $$ We can now multiply both sides by $dt$ again, and get: $$dx = gt\,dt + C\,dt \\ \int dx = \int gt\,dt + C\,dt \\ x = \frac{1}{2}gt^2 + Ct + D $$

Answer 2

$$\dfrac{d^2x}{dt^2}=g$$

$$\dfrac{dx}{dt}=\int g \text{ } dt +C$$

$$x=\int \left[ \int g \text{ } dt +C \right] dt=\iint g \text{ } dt\text{ }dt + \int C dt$$