ODE with three constants

Question

How do we find a third order ODE for harmonic motion with arbitrary phase difference, amplitude and circular frequency? Simple as it seems elimination of non-zero $c_3$ is unclear. Thanks.

$$ y = c_1 \cos( c_2 t + c_3)?$$

Answer 1

The solution can be rewritten as

$$ y = y_0\cos(\omega t + \phi) = A\cos(\omega t) + B\sin(\omega t) $$

where $ A = y_0\cos(\phi) $ and $ B = -y_0\sin(\phi) $. With fixed $\omega$, the above is a general solution of

$$ y'' + \omega^2 y = 0 $$

To cancel $\omega$, rearrange and differentiate

$$ \frac{y''}{y} = -\omega^2 \implies \left(\frac{y''}{y}\right)' = 0 \implies yy''' - y'y'' = 0 $$

Note that this equation may also have hyperbolic solutions, so you'll need to pick initial conditions such that $\frac{y''(0)}{y(0)} < 0$