Given data in the problem
- ${\psi'(t)}_{3 \times 3}=A_{3 \times 3}\psi(t)_{3 \times 3}, \psi(0)_{3 \times 3}=R^{cl}_{3 \times 3} \\ \phi'(t)_{3 \times 3}=t\hspace{.1cm}B_{3 \times 3} \phi(t)_{3 \times 3},\phi(0)_{3 \times 3}=R^{cl}_{3 \times 3} \tag 1$
- $A,B ,R^{cl}$ are constant matrices. A,B are skew symmetric matrices. $R^{cl}$ is a rotation matrix
- We know the solutions of equation (1). Implies we know about what is $\psi(t)=e^{At}R^{cl}_{3 \times 3},\phi(t)=e^{B\frac{t^2}{2}}R^{cl}_{3 \times 3} \tag 2 $
Question
What is the the solution of $ R'(t)=AR(t)+tBR(t)\tag 3$ in closed form?