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

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

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, $\theta_1'=\omega_1,\theta_2'=\omega_2$

  • $\begingroup$ Please consider formatting your math using these instructions to make it easier to read. $\endgroup$
    – user856
    Mar 11, 2013 at 8:20
  • $\begingroup$ To solve a second order DE you reduce it to a pair of two first order DEs. To solve a pair of second order DEs you should reduce to a system of four first order DEs. $\endgroup$ Mar 11, 2013 at 8:56
  • $\begingroup$ How can i do that? $\endgroup$
    – user41235
    Mar 11, 2013 at 13:20
  • $\begingroup$ 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. $\endgroup$ Mar 11, 2013 at 13:35
  • $\begingroup$ see this link. your equation is similar to it. math.stackexchange.com/questions/146823/… $\endgroup$
    – user116132
    Dec 16, 2013 at 12:15

1 Answer 1


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.


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