Following the hint, we have
d V 1 d t = − i 1 C 1 = I / 3 − V 1 / 6 + V 2 / 6 d V 2 d t = i 2 C 2 = I / 6 + V 1 / 6 − V 2 / 6 d I d t = − V L = − 110 I / 3 − 20 V 1 / 3 − 10 V 2 / 3 . {\displaystyle {\begin{aligned}{\frac {dV_{1}}{dt}}&=&-{\frac {i_{1}}{C_{1}}}&=&I/3-V_{1}/6+V_{2}/6\\{\frac {dV_{2}}{dt}}&=&{\frac {i_{2}}{C_{2}}}&=&I/6+V_{1}/6-V_{2}/6\\{\frac {dI}{dt}}&=&-{\frac {V}{L}}&=&-110I/3-20V_{1}/3-10V_{2}/3\end{aligned}}.}
Answer:
[ d V 1 d t d V 2 d t d I d t ] = [ 1 / 3 − 1 / 6 1 / 6 1 / 6 1 / 6 − 1 / 6 − 110 / 3 − 20 / 3 − 10 / 3 ] [ I V 1 V 2 ] {\displaystyle \color {blue}{\begin{bmatrix}{\frac {dV_{1}}{dt}}\\{\frac {dV_{2}}{dt}}\\{\frac {dI}{dt}}\end{bmatrix}}={\begin{bmatrix}1/3&-1/6&1/6\\1/6&1/6&-1/6\\-110/3&-20/3&-10/3\end{bmatrix}}{\begin{bmatrix}I\\V_{1}\\V_{2}\end{bmatrix}}}
