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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 $\... 1 You substitute$\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}$. ... 0 Because you take the derivative with respect to$\tau$. Since$\tau\$ is dimensionless, the derivative is too.