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).




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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,
$$x(t) = \left[\begin{array}{c}
 \sum\_{i=1}^n c\_iv\_i\cos\left(\sqrt{\lambda\_i} t\right) \\ 
 -\sum\_{i=1}^n c\_iv\_i\sqrt{\lambda\_i}\sin\left(\sqrt{\lambda\_i} t\right)
 \end{array}\right].$$



Solve a linear matrix ODE involving symmetric blocks
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 $x(t)={{O^T\; 0}\choose{0\; O^T}} X(t)$.


