= Solution
The maximum of $E(t)/E(0)$ is the square of the largest <singular value> of $A(t)$, hence the largest eigenvalue $\lambda$ of $A(t)^tA(t)$. Its determinant is one and its trace gives
$$
\boxed{(\lambda-1)^2
=\frac{4\beta^2}{1-\beta^2}\lambda\sin^2\Gamma t}.
$$
For $0<\beta<1$, this is largest when $|\sin\Gamma t|=1$, namely $\Gamma t=\pi/2$ modulo $\pi$. Then
$$
\boxed{\lambda_{\max}=\frac{1+\beta}{1-\beta}}.
$$
At such a time $A(t)$ is off diagonal, and the maximizing initial condition is $\boxed{x(0)=0}$ with $y(0)\ne0$. For negative $\beta$, the axes interchange and the formula uses $|\beta|$.
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