I am trying to determine how to turn these two 2nd order ODEs into a system of four 1st order ODEs with initial conditions. Afterwards, I am to turn it into a matrix.
eqn 1: $$3.2\theta''_1 + 1.44[\cos(\theta_2-\theta_1)]\theta''_2 - 1.44[\sin(\theta_2-\theta_1)]\theta'^2_2+62.784\sin(\theta_1)=0$$
eqn 2: $$1.08\theta''_2+1.44[\cos(\theta_2-\theta_1)]\theta''_1+1.44[\sin(\theta_2-\theta_1)\theta'^2_1+17.658\sin(\theta_2)=0$$
Initial Conditions: $$\theta_1(0)=0.2 rad$$ $$\theta'_1(0)=0.4 rad/s$$ $$\theta_2(0)=0.3 rad$$ $$\theta'_2(0)=0.5 rad/s$$
I have tried doing some variable substitution, but it has not gotten me anywhere. What stumps me the most is the variables being embedded within a sine function $\sin(\theta)$ and the $\theta'^2$ as I have never come across it before in my diff eq classes. Does this also require some sort of Laplace Transform?
You can write your ODE in the following form: $$ \begin{pmatrix} 3.2 & 1.44[\cos(\theta_2-\theta_1)] \\ 1.44[\cos(\theta_2-\theta_1)] & 1.08 \end{pmatrix} \vec \theta'' = \vec f(\vec\theta,\vec\theta'), $$ where $f(\ldots)$ encompasses all of the other terms you show.
If you denote the matrix by $A(\vec\theta)$, then you can multiply through with $A^{-1}$ to obtain $$ \vec \theta'' = A(\vec\theta)^{-1} \vec f(\vec\theta,\vec\theta'), $$ and now you can probably see where this is going: If you introduce $\vec\psi=\vec\theta'$, you can rewrite the differential equation in first order form as $$ \vec\theta' = \vec\psi \\ \vec \psi' = A(\vec\theta)^{-1} \vec f(\vec\theta,\vec\psi). $$ This is the desired first order form.