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

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

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

Thomas Russell
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user41235
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  • Please consider formatting your math using these instructions to make it easier to read. –  Mar 11 '13 at 08:20
  • 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. – in_mathematica_we_trust Mar 11 '13 at 08:56
  • How can i do that? – user41235 Mar 11 '13 at 13:20
  • 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. – in_mathematica_we_trust Mar 11 '13 at 13:35
  • see this link. your equation is similar to it.

    http://math.stackexchange.com/questions/146823/solving-coupled-2nd-order-odes-with-runge-kutta-4

     –  Dec 16 '13 at 12:15
  • @Shaktal: I've removed the "coupled-ode" tag. For future reference for systems of ordinary differential equations, tag both as [tag:differential-equations] and [tag:systems-of-equations]. – Willie Wong Feb 13 '14 at 11:05

1 Answers1

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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.

Ross Millikan
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