Differentiating an ODE Suppose we have the ODE $\dot{u}^2 = au^2 + bu +c $ with real constants $a,b,c$ and initial values $u(0) = \dot{u} (0) = 0$. Since I know no better way to solve this, I want to differentiate this ODE, so I get
$$2\dot{u}\ddot{u} = 2au \dot{u} + b \dot{u}.$$
After cancelling $\dot{u}$ we have a linear second-order ODE which is no problem to solve, but this gives us our required $u$ only up to a constant. Is there a fast / easy way to get this constant? If I plug the linear-ODE-solution into the equation from the beginning, the calculation gets really annoying.
 A: Note first that a first order ODE only has one initial condition. With two initial conditions you get a condition on the coefficients on the right side, for instance $$c=\dot u(0)^2-au(0)^2-bu(0).$$

By your equation, the points lie on a quadratic curve, either an ellipse for $a<0$ or a hyperbolic curve for $a>0$. In every case you can achieve a "normal form" with $b=0$ with a constant  shift of $u$ by setting $\tilde u=u+\frac{b}{2a}$, for which you would get the equation $$\dot{\tilde u}^2=a\tilde u^2+c-\frac{b^2}{4a}.$$ 

Renaming the variables we can consider the reduced or normalized case
$$
\dot u^2=au^2+c
$$
Depending on the signs of the coefficients, a different parametrization of the quadratic curve must be chosen. This then results in a simplified ODE.


*

*Case $a=-w^2<0$. Then necessarily $c=r^2>0$. Parametrize $$\dot u^2+(wu)^2=r^2$$ with trigonometric functions, $wu(t)=r\sin v(t)$, $\dot u(t)=r\cos v(t)$. This parametrization has to be compatible with the differential relation, thus for $\cos v(t)\ne 0$ $$wr\cos v=w\dot u=r\cos(v)\dot v\implies \dot v=w.$$ So you get a solution $u(t)=\frac rw\sin(wt+\phi)$. At points $\cos v(t)=0$ the solution can switch to stay stationary for some time.

*Case $a=w^2>0$, $c=r^2>0$. This repeats the above method, only with hyperbolic instead of trigonometric functions. $u(t)=\frac{r}{w}\sinh(wt+ϕ)$.

*Case $a=w^2>0$, $c=-r^2<0$. Now $r^2=w^2u^2-\dot u^2$ has to be parametrized as $wu=r\cosh v$, $\dot u=r\sinh v$ leading to branching points whenever $v=0$, else $$wr\sinh v=w\dot u=r\sinh(v)\dot v$$ so that one concludes again $u(t)=\frac{r}{w}\cosh(wt+ϕ)$.
