Solving ODE containing matrices

Question

We have an ODE $ \psi'(t)_{_{3 \times 3}}=\psi(t)_{3 \times 3}(A_{3 \times 3}+B_{3 \times 3}t)\tag 1$

Given Data in Question

1. We have no quarentee that $\psi'(t),\psi(t)$ both have inverse
2. A,B are skew symmetric constant matrices with determinant $0$
3. We know what is $\psi(0),\psi'(0)$

Question

1. How do we solve this ODE and find out what is $\psi(t)$? Means looking for a closed form.

Answer 1

This is matrix equation that can be written as three identical vector equation: $z'(t)_{1\times3} = z(t)_{1\times3}(A_{3\times3}+B_{3\times3}t)$
Where z is one of the rows of $\psi(t)_{3\times3}$

The solution is of the form: $z=\vec{v}_1e^t+\vec{v}_2e^{t^2}$
Substitute in original equation: $v_1e^t+2tv_2e^{t^2}=(v_1e^t+v_2e^{t^2})(A+Bt)$

$e^t: v_1=v_1A$
$te^t: 0=v_1B$
$e^{t^2}: 0=v_2A$
$te^{t^2}: 2v_2=v_2B$

$v_1\in Null(B)\bigcap Null(A-I)$
$v_2\in Null(A)\bigcap Null(B-2I)$

And now the initial condition can be used to find specific $v_1,v_2$

Repeat this proccess three times to obtain $\psi(t)$

Answer 2

ATTEMPT #1

1. Take transpose on both sides,then you get

   $ (\psi^T(t)_{_{3 \times 3}})^{'}=-(A_{3 \times 3}+B_{3 \times 3}t) \psi(t)_{3 \times 3}^{T}\tag 1$ We know $-(A_{3 \times 3}+B_{3 \times 3}t) $ is a skew symmetric matrix of order $3\times 3$

2. Now we use a theorem given as follows If we have a angular velocity vector $w(t)$,then we can make a rotation matrix of $3\times 3 $ out of it as follows $R(t)=cos(|w(t)|)I+ sin(|w(t)|) [\frac{w(t)}{|w(t)|}]_{\times}+ (1-Cos(|w(t)|)) \left(\frac{w(t)}{|w(t)|}\otimes\frac{w(t)}{|w(t)|}\right)$.

3. And we know that derivative of any rotation matrix makes the relationship with angular velocity as follows $R'(t)=[w(t)]_{\times}R(t) \tag 3$ this analogues to equation (1). Now substitute $[w(t)]_{\times}=-(A_{3 \times 3}+B_{3 \times 3}t)$ and extract $w(t)$ from $-(A_{3 \times 3}+B_{3 \times 3}t)$ ,get R(t). Then $R(t)^T=\psi(t)$. Hope it has solved the issue.. Please comment