# How to solve this coupled 2nd order Differential equation of a double pendulum- Runge Kutta method

$$\theta_1 ''=\frac{-g(2m_1+m_2)sin\theta_1-m_2gsin(\theta_1-2\theta_2)-2sin(\theta_1-\theta_2)m_2(\theta_2'^2l_2+\theta_1'^2l_1cos(\theta_1-\theta_2)}{l_1(2m_1+m_2-m_2cos(2cos(2\theta_1-2\theta_2)))}$$

$$\theta_2 ''=\frac{2sin(\theta_1-\theta_2)(\theta_1'l_1(m_1+m_2)+g(m_1+m_2)cos\theta_1+\theta_2'^2l_2m_2cos(\theta_1-\theta_2))}{l_2(2m_1+m_2-m_2cos(2cos(2\theta_1-2\theta_2)))}$$

These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method?

$m_1,m_2=$masses of pendulum 1 and 2, $\theta_1,\theta_2=$ angles formed by the pendulums, $\theta_1'=\omega_1,\theta_2'=\omega_2$

• You introduce new variables $\omega_{1}$ and $\omega_{2}$ (as you have done) and then you have $\theta_{1}' = \omega_{1}$, $\theta_{2}' = \omega_{2}$, $\omega_{1}' = \theta_{1}'' = ...$ and $\omega_{2}' = \theta_{2}'' = ...$ where you can fill in the blanks yourself. Mar 11, 2013 at 13:35
You view your equations as a four-dimensional vector, (\theta_1,theta_2,\omega_1,\omega_2)^T. You have four first order differential equations, the two big ones you wrote are $\omega_1',\omega_2'$ and your later $\theta_1'=\omega_1,\theta_2'=\omega_2$. Now you are in the form $y'=f(y)$ so you can calculate the intermediate terms just fine.