I've been reading a book on orbit determination, and I've hit a block with the calculus. Why do this?
He defines tau, called modified time, as follows:
$\tau \equiv k(t - t_0)$
Where $k$ is the Gaussian gravitational constant, $t$ is a given instant of time, and $t_0$ is an arbitrarily chosen initial time or epoch.
And then presents the differential of $\tau$ $d \tau = kdt$
The author previously derives an equation from Newton's law of universal gravitation
$\vec a = - { k^2 \mu \vec r \over r^3 } + \sum_{q=1}^{n} k^2 m_q ( {\vec p_q \over p_q^3 } - {\vec r_q \over r_q^3} ),$
Where $\mu = 1 + m_2$ and $m_2$ is in units of whatever the central mass is.
He differentiates the derived equation with respect to modified time...
$\vec {\dot {r}} = ( \frac 1 k) \vec v$
$\vec {\ddot {r}} = ({ 1 \over k^2 }) \vec a$
And we're left with $\vec { \ddot{r}} = -{\mu \vec r \over r^3 } + \sum_{q=1}^{n} m_q ( {\vec p_q \over p_q^3 } - {\vec r_q \over r_q^3} ),$
So, he kills off the $k^2$. He remarks that in principle this simplified equation can be integrated to yield the body's position and velocity at some other time.
There's a lot of good questions here.
We start with $\tau = k(t - t_0)$, a sort of modification of time. You first ask about the differential of $\tau$. While it is possible (and often useful) to think of or consider differentials, let's present the same material without separating $d\tau$ and $d t$ [for a good reference on differentials, consider reading this page, especially examples 2 and 3]. With the definition of $\tau$ above, we see that $$ \frac{d t}{d\tau} = \frac{1}{k}.$$ You might notice that this is a "rearrangement" of the "differential" phrasing you gave above.
With this and the chain rule of calculus, we can write down the claimed versions of velocity and acceleration with respect to this new modification of time. We have that $$ \dot{r} = \frac{dx}{d\tau} = \frac{dx}{dt} \frac{dt}{d\tau} = v \frac{1}{k},$$ where the middle equality is the chain rule in action. Similarly, $$ \ddot{r} = \frac{d \dot r}{d\tau} = \frac{d\frac{v}{k}}{d\tau} = \frac{1}{k} \frac{dv}{dt}\frac{dt}{d\tau} = \frac{1}{k^2} a.$$
So this brings us all the way to the statement of the final equation: $$\vec { \ddot{r}} = -{\mu \vec r \over r^3 } + \sum_{q=1}^{n} m_q \left( {\vec p_q \over p_q^3 } - {\vec r_q \over r_q^3} \right).$$
The author is correct to say that this can be integrated in principle. He's also right to defer it to later, since this is a differential equation that should be done numerically. Without knowing the book (or even the subject), I can't say much more about your author's plans. But it is not unreasonable to go and try to look at how one might numerically solve that differential equation. (Or you could always ask another question here).