Coupled differential equations, no idea on how to solve it due to the constant, university maths

As you haven't shown any work I don't know how to best help you.

Here's a sketch:
Write the system as
[math] \dfrac{d}{dt} \left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right ) = - \omega \left ( \begin{matrix} x_2 \\ x_1 \end{matrix} \right ) = M \left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right )[/math]
So step 1: Find M.

Similar to [math]x' = mx[/math] we are going to look for solutions of the form [math]x(t) = a e^{m t}[/math].

or specifically (skipping a bunch of steps in the derivation)
[math]\left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right ) = \left ( \begin{matrix} a \\ b \end{matrix} \right ) _1 e^{ \lambda _1 t} + \left ( \begin{matrix} a \\ b \end{matrix} \right ) _2 e^{ \lambda _2 t} [/math]
where [math]\lambda _i[/math] and [math]\left ( \begin{matrix} a \\ b \end{matrix} \right ) _i[/math] are the ith eigenvalue and eigenvector of M.

Step 2: Find the eigenvalues and eigenvectors of M.

See what you can do with this.

-Dan
 
Thank you for the help, I had a brain fart and got one of the equations correct and I didnt use the boundary conditions which im quite baffeled, could you check my working out again thanks,
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You were okay until you tried to solve the quadratic.

How do you solve [math]\lambda ^2 - \omega ^2 = 0[/math]? C'mon! You don't need the quadratic formula here. [math]\lambda = \pm \omega[/math] by inspection.

Now continue.

-Dan
 
You were okay until you tried to solve the quadratic.

How do you solve [math]\lambda ^2 - \omega ^2 = 0[/math]? C'mon! You don't need the quadratic formula here. [math]\lambda = \pm \omega[/math] by inspection.

Now continue.

-Dan

Sorry Dan was being a bit of a eejet,
still getting one of the equations wrong at the end ?
 

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