# How to solve $\ddot{\theta} + \mu \dot{\theta}^2=0$

$\dot{\theta} \equiv \frac{d \theta(t)}{dt}$

I encountered this ODE

$\ddot{\theta} + \mu \dot{\theta}^2=0$

How do I find a solution for $\theta(t)$. I know $\dot{\theta}(t=0)=\omega_0$

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Let $u = \dot{\theta}$. The equation now becomes

$\dot{u}+mu^2=0$

This can now simply be solved as a straightforward separable differential equation.

$\dot{u}=-mu^2 \rightarrow \frac{\dot{u}}{u^2}=-m$

Which yields

$\frac{1}{u}=mt+C \rightarrow u=\frac{1}{mt+C}$

$\dot{\theta}$ is now an explicit function of t, so $\theta$ can be integrated directly (with your boundary conditions, so $C$ in this case is $\frac{1}{\omega_0}$).

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Let $y = \dot{\theta}$, then we get $\dot{y} + \mu y^2 = 0$. Assuming $y \neq 0$, we get $\frac{dy}{y^2} + \mu dt = 0 \Rightarrow -\frac{1}{y(t)} + \frac{1}{y(0)} + \mu t = 0 \Rightarrow y(t) = \frac{1}{\mu t + \frac{1}{y(0)}}$
Hence, we have $\frac{d \theta}{dt} = \frac{1}{\mu t + \frac{1}{y(0)}} \Rightarrow \theta(t) = \frac{1}{\mu} \log(\mu t+ \frac{1}{y(0)}) + \theta(0) = \frac{1}{\mu} \log(\mu t+ \frac{1}{\omega_0}) + \theta(0)$