Why does the ODE $$ \theta^{\prime} = (1-\theta^2)^{3/2}, \quad \theta(0) = 0, $$
have the two (different) solutions
$$\theta_1(t) = \frac{t}{\sqrt{1+t^{2}}}, \quad \theta_2(t) = \frac{-t}{\sqrt{1+t^{2}}}, $$
for $t \in (-\delta, \delta)$, whereby $0<\delta <1$?
How is this possible, since the right-hand side of the ODE is locally Lipschitz in a small neighbourhood around $0$ ?
Please make the universe work again.
$\theta_1$ is indeed a solution of the ODE, but $\theta_2$ is not.
$$\theta_2^\prime(t) = -\frac{1}{(1+t^2)^{3/2}} \neq \frac{1}{(1+t^2)^{3/2}} =\left(1-\theta_2^2(t)\right)^{3/2}$$