Solve $ -mg + c \left( \frac{dx}{dt} \right)^2 = m \frac{d^2x}{dt^2} $ Hi we are learning some mathematical modelling processes and producing models for examples such as a skydiver falling out of a plane.
I formed the Differential Equation: $$ -mg + c \left( \frac{dx}{dt} \right)^2 = m \frac{d^2x}{dt^2} $$
I was wondering if this is solvable for $x(t)$ with initial conditions of $\frac{dx}{dt}(0)=0$ and $x(0)=0$?
 A: Let $v\equiv \frac{dx}{dt}$
$$ m \frac{dv}{dt}= cv^2-mg  $$
$$\int dt = \int \frac{mdv}{cv^2-mg}=\frac mc\int \frac{dv}{v^2-\frac{mg}c} $$
Now define the terminal velocity by $v_T \equiv \sqrt\frac{mg}c$ so that ...
$$g\int dt = v_T^2\int \frac{dv}{v^2-v_T^2} $$
$$\implies gt+C= \frac{v_T}2 \ln \left|\frac{v-v_T}{v+v_T}\right |$$
Putting in the absolute value signs allows us to see that $v(0)=0 \implies C=0$
The equation is describing a particle falling from rest under the influence of a constant gravitational field and air resistance, so we expect $v\in [0,-v_T)$
So $\left|\frac{v-v_T}{v+v_T}\right | = -\frac{v-v_T}{v+v_T} = - \left(  1-\frac{2v_T}{v+v_T}\right)$
Rearranging and exponentiating both sides ...
$$ - \left(  1-\frac{2v_T}{v+v_T}\right)=e^\frac{2gt}{v_T} $$
Solving for $v(t)$ ...
$$ v(t) = \frac{2v_T}{1+e^\frac{2gt}{v_T} }  -v_T$$
Integrate to get $x(t)$
$$x(t)=2v_T(-\frac{v_T}{2g}\ln(1+e^{-\frac{2gt}{v_T}}))-v_Tt +C $$
$x(0)=0 \implies C = \frac{v_T^2}{g}\ln2$
The final result can then be written ...
$$x(t)=\frac{v_T^2}{g}(\ln 2 - \ln(1+e^{-\frac{2gt}{v_T}}))-v_Tt $$
