So I have gotten this far... $a = \left(\left({F}_{a} - \frac{{c}_{d} \cdot {\rho}_{h} \cdot A \cdot {v}^{2}}{2}\right) - {m}_{t} \cdot {g}_{r}\right)$ This is based upon ${F}_{a} - \frac{{F}_{d} + {F}_{g}}{m} = a$ But $a$ is $\frac{\mathrm{dv}}{\mathrm{dt}}$ So this becomes a differential equation because the faster you go the harder it is to go faster. How can one solve this? Please note I'm currently a Calc AB student.

Jun 7, 2016

Supposing that
${c}_{0} = {F}_{a} - {m}_{t} {g}_{r} = {C}^{t e}$
${c}_{2} = \frac{{c}_{d} {\rho}_{h} A}{2} = {C}^{t e}$
we have v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] ( t+t_0)) with ${t}_{0}$ integration constant

Explanation:

This equation

$\dot{v} \left(t\right) = {c}_{0} \left(t\right) + {c}_{1} \left(t\right) v \left(t\right) + {c}_{2} \left(t\right) v {\left(t\right)}^{2}$

with $V = {c}_{2} \left(t\right) v$, $S \left(t\right) = {c}_{0} \left(t\right) {c}_{2} \left(t\right)$ and $R \left(t\right) = {c}_{1} \left(t\right) + \frac{{\dot{c}}_{2} \left(t\right)}{{c}_{2} \left(t\right)}$

can be reduced to

$\dot{V} = {V}^{2} + R \left(t\right) V + S \left(t\right)$

which is the known Riccati equation.
https://en.wikipedia.org/wiki/Riccati_equation

Now making $V = - \frac{\dot{u}}{u}$ the Riccati equation can be reduced to
a second order ODE which is

$\ddot{u} - R \left(t\right) \dot{u} + S \left(t\right) u = 0$

A solution $u$ to this equation can be used to find the solution to the original equation

$v = - \frac{\dot{u}}{{c}_{2} \left(t\right) u}$

As an example, supposing that

${c}_{0} = {F}_{a} - {m}_{t} {g}_{r} = {C}^{t e}$
${c}_{2} = \frac{{c}_{d} {\rho}_{h} A}{2} = {C}^{t e}$

we obtain

v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] (t + t_0))

after solving

$\ddot{u} - {c}_{0} {c}_{2} u = 0$

with general solution

$u \left(t\right) = {C}_{1} {e}^{\sqrt{{c}_{0} {c}_{2}} t} + {C}_{2} {e}^{- \sqrt{{c}_{0} {c}_{2}} t}$

Explanation . substituting in

$v \left(t\right) = - \frac{\dot{u} \left(t\right)}{{c}_{2} \left(t\right) u \left(t\right)}$

we get

v(t) = (sqrt[c_0 c_2] C_2 e^(-sqrt[c_0 c_2] t) - C_1 sqrt[c_0 c_2] e^( sqrt[c_0 c_2] t))/(c_2 (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t)))

grouping

v(t) = sqrt(c_0c_2)/c_2((C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t)))

or

v(t) = sqrt(c_0c_2)/c_2((C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t)))

but

(C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t))=(e^(-sqrt[c_0 c_2] t) - C_1/C_2e^(sqrt[c_0 c_2] t))/ (e^(-sqrt[c_0 c_2] t) + C_1/C_2 e^(sqrt[c_0 c_2] t))

and $\exists a | \left\mid {C}_{1} / {C}_{2} \right\mid = {e}^{2 a}$ then

(C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t))=(e^(-sqrt[c_0 c_2] t-a) - e^(sqrt[c_0 c_2] t+a))/ (e^(-sqrt[c_0 c_2] t-a) + e^(sqrt[c_0 c_2] t+a)) = sinh(sqrt(c_0c_2t+a))/cosh(sqrt(c_0c_2t+a)) = tanh(sqrt(c_0c_2t+a))

Finally putting all together

$v \left(t\right) = \sqrt{{c}_{0} / {c}_{2}} T a n h \left(\sqrt{{c}_{0} {c}_{2}} t + a\right)$ or

v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] (t + t_0))

with $a = {t}_{0} \sqrt{{c}_{0} {c}_{2}}$