Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I encountered this ODE

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

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

share|improve this question

3 Answers 3

up vote 3 down vote accepted

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 $\frac{1}{\omega_0}$).

share|improve this answer

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 $\frac{d \theta}{dt} = \frac{1}{\mu t + \frac{1}{y(0)}} \Rightarrow \theta(t) = \frac{1}{\mu} \log(\mu t+ \frac{1}{y(0)}) + \theta(0) = \frac{1}{\mu} \log(\mu t+ \frac{1}{\omega_0}) + \theta(0)$

share|improve this answer
    
Thank you. How do I give two correct answers? –  Please Delete Account Feb 27 '11 at 8:59
1  
You cannot. No problem. –  user17762 Feb 27 '11 at 9:00

This is a Riccati equation :

http://en.wikipedia.org/wiki/Riccati_equation

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.