I have the following rational model equation which fits distance versus time data very well:
$$y = \frac{bt}{1+ct+dt^2}$$
The derivative of this equation is :
$$\frac{dy}{dt} = \frac{(b-bdt^2)}{(1+ct+dt^2)^2}$$
Now at $t = 0, y = 0$ and $$\frac{dy}{dt} =u$$
where $u$ is the velocity at $t=0$
At $t = t_1, y = r$ and $$\frac{dy}{dt} =v$$
what I would like to do is solve for the constants $b$, $c$ and $d$ in terms of $u$, $v$ and $r$:
I can easily get:
$$b = u$$
but I'm having trouble solving for the other coefficients. Is it possible to solve for $c$ and $d$ in terms of $u$, $v$ and $r$?
Let $\Delta := 1+ct_1+dt_1^2$.
At time $t_1$ you have $y = r$ so $\frac{b t_1}{ \Delta} = r$ and $\dot{y} = v$ so $\frac{b(1-dt_1^2)}{\Delta^2} = v$.
You already solved $b = u$. Hence, the first equation yields $\Delta = \frac{ut_1}{r}$. Now $\Delta$ is thus known.
Substituting in the second leads to $b(1-dt_1^2) = v \Delta^2$, from which you can determine $d$.
Eventually, going back to the definition of $\Delta$ and knowing $d$, you can compute $c$.