Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

Following the hint, we have

$\begin{array}{rlrlr}\frac{d{V}_1}{dt}& =& -\frac{i_1}{C_1}& =& I/3-V_1/6+V_2/6\\ \frac{d{V}_2}{dt}& =& \frac{i_2}{C_2}& =& I/6+V_1/6-V_2/6\\ \frac{dI}{dt}& =& -\frac{V}{L}& =& -110I/3-20V_1/3-10V_2/3\end{array}.$

Answer:

$\left[\begin{array}{c}\frac{d{V}_1}{dt}\\ \frac{d{V}_2}{dt}\\ \frac{dI}{dt}\end{array}\right]=\left[\begin{array}{ccc}1/3& -1/6& 1/6\\ 1/6& 1/6& -1/6\\ -110/3& -20/3& -10/3\end{array}\right]\left[\begin{array}{c}I\\ V_1\\ V_2\end{array}\right]$