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**Aryth** I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

$\displaystyle \frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta$

Where l is length of the pendulum.

Doing some research, I keep finding that the solution is:

$\displaystyle \theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)$

Where $\displaystyle \theta_0$ is the initial condition at time t=0

Now, here's my attempt at a solution:

Since it is linear and homogeneous, we can assume a solution:

$\displaystyle \theta = e^{kt}$

Where k is an arbitrary constant.

This leads to the indicial equation:

$\displaystyle k^2 + \frac{g}{l} = 0$

The solution to which is:

$\displaystyle k = \pm i\sqrt{\frac{g}{l}}$

This leads to two solutions:

$\displaystyle \theta_1 = e^{i\sqrt{\frac{g}{l}}t}$

$\displaystyle \theta_2 = e^{-i\sqrt{\frac{g}{l}}t}$

A linear combination of these leads to the general homogeneous solution:

$\displaystyle \theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}$

Expanding the exponentials into their trigonometric form, we get:

$\displaystyle \theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right]$ $\displaystyle + B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]$

$\displaystyle \theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)$

From here I'm not sure how to proceed...