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As you have observed $\beta(t)=0$, $\forall t$ is a solution to the differential equation with initial condition $\beta(0) =0$. There is a potential problem with uniqueness though: Let's set $A=1$ for simplicity. Then, close to $\beta=0$, your equation looks like $$ \frac {d\beta}{dt}= {\sqrt \beta } \quad (\star). $$ (the $B$ term is quadratic in $\...


You substitute $\dot{\phi} = \frac{1}{T} \frac{d\phi}{d\tau}$. Now let's consider the dimensions \begin{align} [\dot{\phi}] &= [\frac{d\phi}{dt}] = \frac{rad}{s}\\ [\dot{\phi}] &= [\frac{1}{T} \frac{d\phi}{d\tau}] = \frac{1}{s} \cdot [\frac{d\phi}{d\tau}] \\ \Rightarrow [\frac{d\phi}{d\tau}] &= rad \end{align} and analog for $\ddot{\phi}$. ...


Because you take the derivative with respect to $\tau$. Since $\tau$ is dimensionless, the derivative is too.

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