Solution of Matrix ODE

Question

Specifications

1. It is given that $ \psi'(s)=(A+Bs)\psi(s)\tag 1$

   where A,B are constant $3 \times 3$ skew symmetric matrices with determinant $0$

2. $\psi(s)$ has determinant $1$ , orthogonal and has dimension $3 \times 3$

3. We know $\psi(0)=I_{3 \times 3 }$. Where $I_{3 \times 3 }$ Unit Matrix

Question

1. Is it true that solution of equation (1) is $e^{(A+Bs)}$ ? If so how do we prove that?

NB:: I couldn't find a general method to solve it. If it was not matrix I know how to do it. But the issue is these are matrices. More over $(A+Bs)$ skew symmetric with order $3$ means it has no inverse( Remember the initial condition. My hint on solution is based on initial condition $I$)

Answer 1

If $A$ and $B$ commute, then you can probably do something exact. Otherwise, you can start with $$ \psi(s)=\psi(0)+\int_{0}^{s}(A+s_1 B)\psi(s_1)\,ds_1 $$ and begin to iterate: $$ \psi(x) = \psi(0)+\int_{0}^{s}(A+s_{1}B)\left[\psi(0)+\int_{0}^{s_{1}}(A+s_{2}B)\psi(s_{2})\,ds_{2}\right]ds_{1} \\ = \psi(0)+sA\psi(0)+\frac{s^{2}}{2}B\psi(0)+\int_{0}^{s}(A+s_{1}B)\int_{0}^{s_{1}}(A+s_{2}B)\psi(s_{2})\,ds_{2}\,ds_{1}. $$ You can get higher-order terms this way that depend on the initial vector $\psi(0)$ and the matrices $A$, $B$. The series will converge.