Solve a linear matrix ODE involving symmetric blocks Let $P \in \mathbb R^{n \times n}$ be a symmetric positive definite matrix with eigenvalues denoted by $\lambda_i$ and corresponding eigenvectors denoted by $v_i$. For each $j \in \{1, 2, 3, 4\}$, let $\alpha_j$ be a non-zero real number. Let $x: [0, \infty) \rightarrow \mathbb R^{2n}$ be a continuous, differentiable function satisfying 
\begin{align*}
    &\frac{d}{dt}x(t) = Ax(t), \\
    &A = \left[\begin{array}{cc}
    \alpha_1P & \alpha_2I \\
    \alpha_3P & \alpha_4I
    \end{array}\right] \in \mathbb R^{2n \times 2n}, \\
    &x(0) = z \in \mathbb R^{2n},
\end{align*}
where $I \in \mathbb R^{n \times n}$ is the identity matrix. What methods can be used to manually obtain the solution $x(t)$ to the above differential equation, expressed in terms of $\lambda_i$ and $v_i$? We may manipulate $z$ for convenience (for example, as done below).

If we don't require the $\alpha_j$ to be non-zero, the case 
$$A = \left[\begin{array}{cc}
    0 & I \\
    -P & 0
    \end{array}\right]$$
gives rise to the following solution. Define $x_1 \in \mathbb R^n$ and $x_2 \in \mathbb R^n$ such that 
$$x = \left[\begin{array}{c}
    x_1 \\
    x_2
    \end{array}\right],$$
and let
$$x_1(0) = \sum_{i=1}^n c_iv_i,$$
where each $c_i$ is a real number. I've been informed that, in this case,
$$x(t) = \left[\begin{array}{c}
    \sum_{i=1}^n c_iv_i\cos\left(\sqrt{\lambda_i} t\right) \\ 
    -\sum_{i=1}^n c_iv_i\sqrt{\lambda_i}\sin\left(\sqrt{\lambda_i} t\right)
    \end{array}\right].$$
 A: Diagonalize $P=O\Lambda O^T$ with an $n\times n$ orthogonal matrix $O$ (containing the eigenvectors $v_i$) and a diagonal matrix $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_n)$. Define $X={{O\; 0}\choose{0\; O}}x$ and $Z={{O\; 0}\choose{0\; O}}z$. Then the differential equation becomes
$$    \frac{d}{dt}X(t) = \left(\begin{array}{cc}
    \alpha_1\Lambda & \alpha_2I \\
    \alpha_3\Lambda & \alpha_4I
    \end{array}\right)X(t), \;\;
    X(0) = Z, $$
with solution
$$X(t)=M(t)Z,\;\;M(t)=\exp\left(\begin{array}{cc}
    \alpha_1\Lambda\, t & \alpha_2I t \\
    \alpha_3\Lambda\, t & \alpha_4I t
    \end{array}\right)$$
$$\Rightarrow M(t)=\left(
\begin{array}{cc}
 e^{\Omega_+t}\cosh\Xi\, t+\Omega_-\Xi^{-1}e^{\Omega_+t}{\sinh \Xi\, t }& {\alpha_2}\Xi^{-1}e^{\Omega_+t}{\sinh \Xi\, t } \\
 {\alpha_3} \Lambda\Xi^{-1}e^{\Omega_+t}{\sinh \Xi\, t } &e^{\Omega_+t}\cosh\Xi \,t- \Omega_-\Xi^{-1}e^{\Omega_+t}{\sinh\Xi\, t } \\
\end{array}\right),$$
where we have defined
$$\Omega_\pm=\tfrac{1}{2}(\alpha_1\Lambda\pm\alpha_4 I),\;\;\Xi=\sqrt{\Omega_-^2+ \alpha_2\alpha_3\Lambda}.$$
From here you recover $x(t)={{O^T\; 0}\choose{0\; O^T}} X(t)$.
