I have the Lagrangian $L=\frac{1}{2}(\dot{x}^2+\dot{y}^2) -7x^2+2xy-\frac{11}{2}y^2$. So the two equations of motion are $\ddot{x}+14x-2y=0$ and $\ddot{y}-2x+11y=0$. Hence the general solution (by using the normal modes method) is
$\vec{x}(t)=\begin{bmatrix} -2 \\ 1 \\ \end{bmatrix}$$\big(\alpha_1\cos(\sqrt{15}t)+\beta_1\sin(\sqrt{15}t)\big) + \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$$\big(\alpha_2\cos(\sqrt{10}t)+\beta_2\sin(\sqrt{10}t)\big)$.
But here I am asked to find the normal modes associated with oscillations about $x=y=0$. I suspect that this can be summaried by initial conditions for $\vec{x}(t)$ and $\vec{v}(t)$ but I don't know which ones.
As the movement is described by
$$ \ddot X = M\cdot X $$
with $M = T\cdot \Lambda\cdot T^{-1}$ we have
$$ T^{-1}\cdot {\ddot X} = \ddot Z = \Lambda\cdot T^{-1}\cdot X = \Lambda\cdot Z $$
so with a change of coordinates we have the equivalent system
$$ \ddot Z = \Lambda\cdot Z\Rightarrow \cases{\ddot z_1 = \lambda_1 z_1\\ \ddot z_2 = \lambda_2 z_2} $$
and the normal/generator modes are
$$ \cases{ z_1 = \alpha_1\cos\sqrt{\lambda_1 t}+\beta_1\sin\sqrt{\lambda_1}t = \rho_1\cos(\sqrt{\lambda_1}t+\phi_1)\\ z_2 = \alpha_2\cos\sqrt{\lambda_2 t}+\beta_2\sin\sqrt{\lambda_2}t = \rho_2\cos(\sqrt{\lambda_2}t+\phi_2)} $$