How can I solve a differential equation of the form $c_1 \ddot{x}^2 + c_2 \ddot{x} + c_3 \dot{x}^2-c_4 =0$? The solution to it will give us the eqs. of motion of a car moved by a constant force due to combustion of fuel at the engine that produces the main torque at the wheels and that is "fighting" against friction and drag. The differential equation comes from the 2nd law of Newton and the torque equations.
In brief, we'd have
\begin{align*}
\sum F &= m(t) \ddot{x}\\
m(t)\ddot{x} &= F_{fuel} -F_f -F_d = F_{fuel} -\mu m(t)g-\kappa \dot{x}^2
\end{align*}
Since there are two unknown functions we want to find out, $m(t)$ and $x(t)$, $\dot{x}(t)$, or $\ddot{x}(t)$, we'll have to find a relation between $m(t)$ and any of these 3. We'll do it with the torque eqs.. We'll consider that the torque in one of the 4 wheels of the car is generated thanks to $\frac{1}{4}F_f$ and $F_{fuel}$:
\begin{align*}
\sum \tau = r \sin{\theta}\sum F &= I\ddot{\varphi}\\
r(F_{fuel}+\frac{1}{4}F_f) &= \frac{1}{2}m_w r^2 \ddot{\varphi}\\
F_{fuel}+\frac{1}{4}\mu m(t)g &= \frac{1}{2}m_w \ddot{x}\\
m(t) &= \frac{2 m_w}{\mu g} \ddot{x} -\frac{4}{\mu g}F_{fuel}
\end{align*}
Now we substitute this expression in the Newton's equation and get the differential equation from the title for $c_1 \equiv \frac{2 m_w}{\mu g}$, $c_2 \equiv 2m_w -\frac{4}{\mu g} F_{fuel}$, $c_3 \equiv \kappa$, and $c_4 \equiv 5 F_{fuel}$:
\begin{align*}
\Bigg[\frac{2 m_w}{\mu g} \ddot{x} -\frac{4}{\mu g}F_{fuel}\Bigg]\ddot{x} &= F_{fuel} -\mu g \Bigg[\frac{2 m_w}{\mu g} \ddot{x} -\frac{4}{\mu g}F_{fuel}\Bigg]-\kappa \dot{x}^2\\
\frac{2 m_w}{\mu g} \ddot{x}^2 -\frac{4}{\mu g}F_{fuel}\ddot{x} &= F_{fuel} -2 m_w\ddot{x} + 4F_{fuel}-\kappa \dot{x}^2\\
\frac{2 m_w}{\mu g} \ddot{x}^2 + (2m_w -\frac{4}{\mu g} F_{fuel})\ddot{x} + \kappa \dot{x}^2 -5 F_{fuel}&= 0
\end{align*}
I don't have the enough knowledge to approach this diff. eq. yet, and I don't even know if it is solvable analitically.
 A: I will concentrate on the solution of the ODE $c_1 \ddot{x}^2 + c_2 \ddot{x} + c_3 \dot{x}^2-c_4 =0$ without comment/reference to the physics in the post. I do not believe that the solution can be obtained explicitly. However, some analytical progress can be made.
Let us set $v = \dot x$. Then we need to solve the first order equation
$$c_1 \dot{v}^2 + c_2 \dot{v} + c_3 v^2-c_4 =0\,. \tag{1}$$
This equation is solvable by separation of variables.
Starting from (1), we obtain the expression
$$ \dot v = - \frac{c_2 \pm \sqrt{c_2^2 + 4 c_1 c_4 -4 c_1 c_2 v}}{2 c_1} $$
for the acceleration. The initial condition will typically tell, which of the two solutions you should take. We can integrate the equation with the result
$$ \int_{v_0}^{v} \frac{dv'}{c_2 \pm \sqrt{c_2^2 + 4 c_1 c_4 -4 c_1 c_2 v'^2}} = - 2 c_1 (t-t_0)$$
which implicitly determines $v(t)$. In particular, the integral can be evaluated in terms of elementary functions.
The remaining step is then to integrate $v(t)$ to obtain the position $x(t)$.
A: Here's the parametric solution. Starting from where Fabian has left off,
\begin{align}\tag{1}
t=t_0-\frac{1}{2c_1}\int_{x'_0}^{x'} \frac{\mathrm dv}{c_2\pm\sqrt{c_2^2+4c_1 c_4-4c_1 c_2v^2}},\quad\longrightarrow \quad t=t_0+F(x').
\end{align}
Taking $\xi=x'$, taking a derivative of equation (1), and substituting $\mathrm dt=\mathrm dx/\xi$ we find that \begin{align}
\mathrm dx&=\xi F(\xi)'\mathrm d\xi,\\
x&=x_0+\int_{x'_0}^\xi sF(s)'\mathrm ds.
\end{align}
This integral comes out a little shorter than the first! I'll leave it to you to sort out the detailsThe parametric solution is then given by
\begin{align}
t=t_0+F(\xi),\quad x=x_0+\int_{x_0'}^{\xi}sF(s)'\mathrm ds.
\end{align}
