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

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To solve the second-order ordinary differential equation (ODE) for a pendulum's displacement when released from rest, the initial condition requires that the displacement at time zero is zero, leading to the relationship A + B = 0, which implies A = -B. The time derivative of the displacement is then evaluated at time zero, resulting in an equation that incorporates the initial velocity v0. This equation simplifies to a form that must hold true for the system, specifically involving the parameters ω1 and Ω. The solution progresses by substituting A = -B into this derived equation to find the constants.
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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