Infinite Series -: $\psi(s)=\psi(0)+\psi_1(0)s+\psi_2(0)\frac{s^2}{2!}+\psi_3(0)\frac{s^3}{3!}+.+.+ $. We have a given converging series using derivatives and matrices(Analogue to Taylor's series)
$\psi(s)_{3 \times 3}=\psi(0)+\psi_1(0)s+\psi_2(0)\frac{s^2}{2!}+\psi_3(0)\frac{s^3}{3!}+..+.. \tag 1$. 
(Note the notional convention used here $\frac{\mathrm{d}^2 \psi(s) }{\mathrm{d} s^2}=\psi_2(s),\frac{\mathrm{d}^p \psi(s) }{\mathrm{d} s^p}=\psi_p(s) $)
Given data and Observation in the question


*

*It is given that $ \psi_2(s)=(A+Bs)\psi_1(s)\tag 2$
where A,B are constant $3 \times 3$ skew symmetric matrices with determinant $0$ 
$$A=\left( \begin{array}{ccc} 
   0 & -c_0 & b_0 \\
   c_0 & 0 & -a_0 \\
  -b_0 & a_0 & 0 \\
   \end{array} \right).$$
$$B=\left( \begin{array}{ccc} 
   0 & -c_1 & b_1 \\
   c_1 & 0 & -a_1\\
  -b_1 & a_1 & 0 \\
   \end{array} \right).$$
Note : All entries of the matrices $A$, $B$ are constants,can't be altered  

*$\psi_1(0)$ has determinant $1$ and  orthogonal. No information about the same property on other derivatives.

*$\psi_1(0),\psi(0)$, are given  
Question


*

*Can we re write equation (1) as $\psi(s)_{3 \times 3}=\tau(s)_{3 \times 3} \psi_1(0)+\psi(0) \tag 3$? 

*Can we re write  $\psi(s)_{3 \times 3}$  in a finite Closed form by summing all the terms? 
NB :: Means what could be the finite function $\tau(s)$ which is defined with out any derivatives of $\psi(s)$. Implies you can write $\tau(s)$ using $A,B,s$ as per your convenience . Please check my attempts to solve it as answers below. If you have different idea you can write new one

ATTEMPT #1#Answer
When you look at the relationship  $ \psi_2(s)=(A+Bs)\psi_1(s) $ we can make recursion out of it 
$\psi(0)_n=A\psi(0)_{n-1}+(n-2)B \hspace{.2cm}\psi(0)_{n-2}\tag 4$
For simplicity let me write $y_n=\psi(0)_n,C_1=A,C_2=B$,because we are going to make recursion on n. All s based variables are now become constant
$y_n =
\left\{
 \begin{array}{ll}
   \psi_1(0)  & \mbox{if } n = 1 \\
   C_1\psi_1(0)& \mbox{if }n=2\\
        C_1y_{n-1}+(n-2)C_2 y_{n-2} & \mbox{if }n> 2\\
 \end{array}
\right.\tag 5$
Question 1
  
  
*
  
*Find out the solution of recursion in equation (5). The solution may  contain initial terms $y_1,y_2$(as most of the recursion solution), then substitute for  each $y_n$. Then you can take out $y_1$ as mentioned in the question.You may need to solve the infinite series after taking out $y_1$. I am trying to solve it. Till not yet successful. Expecting suggestions
  
  
  Question 2
Multiply by $x^n$ ,then take summation
$\sum_{n=1}^{\infty}y_nx^{n}= xC_1\sum_{n=2}^{\infty}y_{n-1}x^{n-1}+x^2C_2\sum_{n=3}^{\infty} (n-2)y_{n-2}x^{n-2} \tag 6  $
$\sum_{n=1}^{\infty}y_nx^n= xC_1\sum_{n=1}^{\infty}y_{n}x^{n}+x^2C_2\sum_{n=1}^{\infty}  n y_{n}x^{n} \tag 7  $
$\sum_{n=1}^{\infty}y_nx^n= xC_1\sum_{n=1}^{\infty}y_{n}x^{n}+x^3C_2\sum_{n=1}^{\infty}  n y_{n}x^{n-1} \tag 8  $
Assume $F(x)_{3 \times 3}=\sum_{n=1}^{\infty}y_nx^n$ then we get an ODE
$ F(x)=xC_1F(x)+x^3C_2F'(x)\tag 9$
Solving $F(x)$ will give solution of question number 2.More precisely $F(1)$. But the issue is I couldnt not find a solution. Expecting suggestions to solve it

 A: Here is some progress towards a solution. To review, we have $\psi:\mathbb{R}\to\mathbb{R}^{3\times 3}$ such that $\psi_0=0$. (Notation: I take $\psi_0^{(n)}:=\frac{d^n\psi}{ds^n}\bigr|_0$). We further assume that $\psi_0'$ is a special orthogonal matrix (determinant 1) and $\psi''(s)=(A+Bs)\psi'(s)$ for constant skew-symmetric matrices $A,B$. (That they have determinant $0$ follows from their skew-symmetry.)
The question first asks if we may find $\tau:\mathbb{R}\to\mathbb{R}^{3\times3}$ such that $\psi(s)=\tau(s)\psi'_0$. This at least is simple: Multiplying both sides by $(\psi_0')^T$ and recalling that $\psi'_0$ is orthogonal, we have $\tau(s)=\psi(s)(\psi'_0)^T.$ So $\tau(s)$ is certainly well-defined.
For the second computational part of the question, we note that inserting the Taylor series $\psi(s)=\sum\limits_{k=1}^\infty \psi_0^{(k)}\dfrac{s^k}{k!}$ yields $\tau(s)=\sum\limits_{k=1}^\infty C_k\dfrac{s^k}{k!}$ where $C_k:=\psi_0^{(k)}(\psi'_0)^T$. The task becomes to characterize $C_k$, or better yet to resum the series explicitly. I will not attempt the latter, and so do not consider this a complete solution.
With an eye towards understanding the $C_k$, observe that $\psi''(s)=(A+Bs)\psi'(s)$ implies $\tau''(s)=(A+Bs)\tau'(s)$. Returning to Taylor series, we have
\begin{align}
\sum\limits_{k=0}^\infty C_{k+2}\dfrac{s^k}{k!}
&=(A+Bs)\sum\limits_{k=0}^\infty C_{k+1}\dfrac{s^k}{k!}\\
&=\sum\limits_{k=0}^\infty AC_{k+1}\dfrac{s^k}{k!}
+\sum\limits_{k=0}^\infty kBC_{k}\dfrac{s^{k}}{k!}.
\end{align}
Inspecting coefficients, we're able to summarize the $C_k$ as  $$C_1=\psi_0'(\psi_0')^T=I, C_{k+2}=AC_{k+1}+kBC_k\: \text{  for } k\geq 0.$$ This is convenient for computation, and in that spirit we list off the first few terms:
$$C_1=I,\;C_2=A,\\ C_3=A^2+B,\\ C_4=A^3+AB+2BA,\\C_5=A^4+A^2B+2ABA+3BA^2+3B^2,\,...$$
Note that the order of these terms is crucial since they need not commute. There is an interesting pattern here, which is most evident in the $C_5$ case: Each term may be generated by starting with $A^{n-1}=AA\ldots A$, replacing consecutive pairs of $A's$ with $B's$, and assigning an overall integer prefactor (whose interpretation isn't obvious to me). It reminds me of certain expressions from Wick's theorem, but an explicit linkage eludes me.
