Is this ODE solvable? this equation popped up when I was trying to apply math:
$$Af^2+B\left(\frac{df}{dx}\right)^2=C$$
Where $A,B,C$ are positive constants. Also,  $~f(0)=D$, another positive constant. What are the solutions? Is this solvable? If not, are there any texts you recommend to read?
BTW you might want to know background: I am solving for the maximum velocity profile $v(s)$ for a car traveling on a curved path $\gamma(s)$ under the constraint that the max passenger g force, call it $\sqrt{C}$, is not exceeded. This results in
$$\sqrt{g_l^2+g_r^2}=\sqrt{C}~~\to~~\left(\frac{v^2}{r}\right)^2+\left(\frac{dv}{dt}\right)^2=C$$
Where $r$ is the radius of curvature at the point $\gamma(s)$. I then made the substitution $f=v^2$ to get to an equation analogous to the one above. Also, I took $r$ to be piecewise (which is a good approximation in my usage case), which is why $A$ is constant.
 A: To solve the equation take the derivative of it:
$$Af^2 + B\left(\frac{df}{dx}\right)^2 = C \implies 2\frac{df}{dx}\left(Af + B \frac{d^2f}{dx^2}\right) = 0$$
So either we have $f = \text{constant}$ or
$$Af + B \frac{d^2f}{dx^2} = 0$$
This is a simple equation to solve ($f(x) = ae^{\omega t} + b ae^{-\omega t}$)  and depending on the sign of $\omega^2 \equiv \frac{A}{B}$ we either get osscillating solutions or exponentially growing/decaying solutions.
A: Maple (software) gives these 4 solutions:
f(x) = sqrt(A*C)/A, 
f(x) = -sqrt(A*C)/A, 
f(x) = sqrt(A*(tan(sqrt(B*A)*(_C1-x)/B)^2+1)*C)*tan(sqrt(B*A)*(_C1-x)/B)/(A*(tan(sqrt(B*A)*(_C1-x)/B)^2+1)), 
f(x) = -sqrt(A*(tan(sqrt(B*A)*(_C1-x)/B)^2+1)*C)*tan(sqrt(B*A)*(_C1-x)/B)/(A*(tan(sqrt(B*A)*(_C1-x)/B)^2+1))
It is solving by straightforward quadrature (integration) without imposing any initial condition. _C1 is an arbitrary constant. If you impose the initial condition $f(0)=D$ then you get only two implicit solutions:
$$x-B\cdot \arctan(\sqrt{B\cdot A}\cdot f(x)/\sqrt{-A\cdot B\cdot f(x)^2+B\cdot C})/\sqrt{B\cdot A}+B\cdot \arctan(\sqrt{B\cdot A}\cdot D/\sqrt{-A\cdot B\cdot D^2+B\cdot C})/\sqrt{B\cdot A} = 0$$
and
$$x+B\cdot \arctan(\sqrt{B\cdot A}\cdot f(x)/\sqrt{-A\cdot B\cdot f(x)^2+B\cdot C})/\sqrt{B\cdot A}-B\cdot \arctan(\sqrt{B\cdot A}\cdot D/\sqrt{-A\cdot B\cdot D^2+B\cdot C})/\sqrt{B\cdot A} = 0$$
