Solve the second order equation $$\frac{d^2u}{dt^2} = \begin{bmatrix} -5 & -1 \\ -1 & -5 \end{bmatrix}u
$$
with
$$
u(0)=\begin{bmatrix}1 \\ 0 \end{bmatrix}
$$
and
$$
u'(0) = \begin{bmatrix}0 \\ 0 \end{bmatrix}
$$
Okay I tried to create a $u$ vector but couldn't. And believe I can't get $\frac{du}{dt}$ by finding the eigenvalues and eigenvectors of the matrix $A$ right?  Since matrix $A$ multiplied by $u$ gives the second derivate, not matrix 
$A$ multiplied by the first derivative.
What to do here?
 A: An equivalent method without (explicitly) using eigenvalues, as the matrix is symmetric:
Writing $u=(u_1,u_2)^T$ the d.e. is equivalent to:
\begin{align}
\frac{d^2 u_1}{d t^2}&=-5u_1-u_2\\
\frac{d^2 u_2}{d t^2}&=-u_1-5u_2\\
\end{align}
Consider instead $u_1+u_2$, and $u_1-u_2$,
\begin{align}
\frac{d^2 }{d t^2}(u_1+u_2)&=-6(u_1+u_2)\\
\frac{d^2 }{d t^2}(u_1-u_2)&=-4(u_1-u_2)\\
\end{align}
Thus $$u_1+u_2=A\cos(t\sqrt{6})+B\sin(t\sqrt{6})$$
and $$u_1-u_2=C\cos(2t)+D\sin(2t)$$
Thus 
\begin{align}
u_1&=\frac{1}{2}(A\cos(t\sqrt{6})+B\sin(t\sqrt{6})+C\cos(2t)+D\sin(2t))\\
u_2&=\frac{1}{2}(A\cos(t\sqrt{6})+B\sin(t\sqrt{6})-C\cos(2t)-D\sin(2t))\\
\end{align}
$u(0)=(1,0)^T \implies \frac{1}{2}(A+C)=1$ and $\frac{1}{2}(A-C)=0$, so $A=C=1$.
$u'(0)=(0, 0)^T \implies \frac{1}{2}(\sqrt{6}B+2D)=0$ and $\frac{1}{2}(\sqrt{6}B-2D)=0$, so $B=D=0$.
Thus the solution is:
$$\begin{pmatrix} u_1\\u_2 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} \cos{t\sqrt{6}}+\cos{2t}\\\cos{t\sqrt{6}}-\cos{2t} \end{pmatrix}$$

Alternatively we can follow the clever answer here by user: Start wearing purple. 
Let $$A=\begin{pmatrix} -5 & -1\\-1 & -5\end{pmatrix}$$
A is symmetric so we can diagonalize it with an orthogonal matrix $P$ whose columns are eigenvectors. The eigenvalues of $A$ are found by solving $\det(\lambda I-A)=0$, which gives $(\lambda+5)^2-1=0$, i.e. $\lambda=-4$ or $\lambda=-6$. Plugging back in lambda and solving for $(A-\lambda I)\vec{x}=0$, gives two eigenvectors $(1, 1)^T$ and $(1, -1)^T$. So take $$P=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}$$
As you can check $P$ is orthogonal, $P^TP=PP^T=I$. Then $$PAP^T=\begin{pmatrix}-6 & 0 \\ 0 & -4\end{pmatrix}=D$$
So if we act on our differential equation by $P^T$ then we have that
$$\frac{d^2}{dt^2}P^T \vec{u}=PA{u}=DP^T\vec{u}$$
Let $\vec{v}=P^T \vec{u}$, then we have the new differential equation:
$$\frac{d^2}{dt^2}\vec{v}=D\vec{v}$$
with new initial conditions $\vec{v}(0)=P^T \vec{u}(0)$ and  $\vec{v}'(0)=P^T \vec{u}'(0)$. This decouples into:
\begin{align}
\frac{d^2 v_1}{d t^2}&=-6v_1\\
\frac{d^2 v_2}{d t^2}&=-4v_2\\
\end{align}
We can solve the way we did before $\textit{or}$ to keep everything in terms of matrices and using that $f(\operatorname{diag}(\lambda_1,\lambda_2))=\operatorname{diag}(f(\lambda_1),f(\lambda_2))$ we can say that:
$$\vec{v}=\cos\left(t\sqrt{\strut-D} \right)\vec{v}(0)+\left(\sqrt{\strut-D}\right)^{-1}\sin\left(t\sqrt{\strut-D} \right)\vec{v}'(0)$$
Using that $P\vec{v}=\vec{u}$ we find that 
\begin{align}
\vec{u}&=P\cos\left(t\sqrt{\strut-D} \right)P^T\vec{u}(0)+P\left(\sqrt{\strut-D}\right)^{-1}\sin\left(t\sqrt{\strut-D} \right)P^T\vec{u}'(0)\\
&=\cos\left(t\sqrt{\strut-A} \right)\vec{u}(0)+\left(\sqrt{\strut-A}\right)^{-1}\sin\left(t\sqrt{\strut-A} \right)\vec{u}'(0)\\
&=\cos\left(t\sqrt{\strut-A} \right)\vec{u}(0)
\end{align}
using that $P^T=P^{-1}$, and $f(PDP^{-1})=f(D)$. Of course it will be easier to evaluate $P\cos\left(t\sqrt{\strut-D} \right)P^T\vec{u}(0)$ which is
\begin{align}
&=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}\cdot\begin{pmatrix}\cos(t\sqrt{6}) & 0 \\ 0 & \cos{2t}\end{pmatrix} \cdot\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}\\
&=\frac{1}{2}\begin{pmatrix} \cos{t\sqrt{6}}+\cos{2t}\\\cos{t\sqrt{6}}-\cos{2t} \end{pmatrix}
\end{align}
