How do you solve this 2nd ODE for a pendulums displacement...

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SUMMARY

The discussion focuses on solving the second-order ordinary differential equation (ODE) for a pendulum's displacement when released from rest with an initial velocity of \( v_0 \). The key equations derived include \( \tilde{r}(0) = A + B = 0 \) leading to \( A = -B \) and the time derivative evaluated at zero, resulting in \( i[(\omega_1 - \Omega)A - B(\Omega + \omega_1)] = v_0 \). The conclusion emphasizes that the equation \( (\omega_1 - \Omega)A - B(\Omega + \omega_1) = -i v_0 \) is crucial for further calculations.

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applestrudle
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..when it is released from rest with velocity (v0, 0)

Screen Shot 2015-09-16 at 15.07.22.png


I can get 1.6.5 but I can't get this:

Screen Shot 2015-09-16 at 15.08.01.png
 
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First, note that ## \tilde{r}(0) = A + B = 0 ##, because the pendulum starts off at the origin. This equation gives ## A = -B ##. Next, calculate the time derivative of ## \tilde{r} ##, evaluate it at time zero, and set it equal to ## v_0 + i(0) = v_0##:

## i[(\omega_1-\Omega)A-B(\Omega+\omega_1)] = v_0 ##.

But the only way this equation could be true is if

## (\omega_1 - \Omega)A - B(\Omega+\omega_1)=-i v_0 ##.

Do you see it? The rest follows by using ## A = -B ##.
 

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