For example, I have the Euler-Lagrange equation for a simple pendulum:
\begin{eqnarray*} \frac{d^2\theta}{dt^2}+\frac{g}{l}\theta &=& 0 \end{eqnarray*}
How do I find $\theta$? I know that one approach is to find the roots of the characteristic equation $\lambda^2+\frac{g}{l}\lambda=0$ and then write the solution in the form $y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x}$, but the roots of such an equation are $0$ and $-\frac{g}{l}$, while the answer seems to entail some sines and cosines. Any ideas?
I'm looking into an ALTERNATIVE SOLUTION:
Two functions will solve the equation in question: $$y(t)=\cos\left(\sqrt \frac{g}{l}t\right)$$ and $$y(t)=\sin\left(\sqrt \frac{g}{l}t\right)$$
The general solution is a linear combination of both sine and cosine, i.e.: $$y(t)=A\cos\left(\sqrt \frac{g}{l}t\right) + B\sin\left(\sqrt \frac{g}{l}t\right)$$
And then this can be rewritten as $$y(t)=A\sin\left(\sqrt \frac{g}{l}t+\theta\right)$$
Can anyone give some intuition into this?
Wouldn't it also make sense to have cosine instead of sine because this is describing general harmonic oscillation path (i.e. $y(t)=Acos(\omega t+\theta)$, where A is the amplitude, $\omega$ is the angular frequency and $\theta$ is a phase change? This equation would then imply that $\sqrt \frac{g}{l}$ is the angular frequency? Is this the right thinking?