Finding normal coordinates of a system We have coupled oscillators with equations of motion: $$\ddot{x} = -10x+18y$$ $$\ddot{y}=-3x+5y$$ At $t= 0$ we have $x=a$ and $\dot{x}=\dot{y}=y=0$. I found the solution to be $$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = a\begin{pmatrix} 3 \\ 1\end{pmatrix}\cos{2t} -a\begin{pmatrix} 2 \\ 1\end{pmatrix}\cos{t}$$
I am given that the system has normal coordinates $u=x-2y$ and $v=x-3y$, but how are these found?

Well, this is only my guess because I have not learnt system of differential equations yet. I hope I'm not mistaken. Your solution can be written as
$$
\begin{align}
\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 &-2\\ 1&-1\end{pmatrix}\begin{pmatrix} u \\ v \end{pmatrix}=A\begin{pmatrix} u \\ v \end{pmatrix}
\end{align}
$$
where $u=a\cos2t$ and $v=a\cos t$. Hence
$$
\begin{align}
\begin{pmatrix} u \\ v \end{pmatrix} &= A^{-1}\begin{pmatrix} x \\ y \end{pmatrix}\\
&=\frac{1}{-1}\begin{pmatrix} -1 &2\\ -1&3\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\\
&=\begin{pmatrix} 1 &-2\\ 1&-3\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\\
\begin{pmatrix} u \\ v \end{pmatrix} &=\begin{pmatrix} x -2y\\ x-3y\end{pmatrix}
\end{align}
$$
I hope this is correct. If not, please do not vote down my answer. Just ignore it. Also, thank you so much for whoever vote up my answer. I really appreciate it. ヾ(-^〇^-)ﾉ
