Science:Math Exam Resources/Courses/MATH152/April 2022/Question B5 (b)/Solution 1

We are given ${\textstyle C_{2}=3F}$ in the question statement, so what remains is to write ${\textstyle J_{2}}$ in terms of ${\textstyle V_{1}}$, ${\textstyle V_{2}}$, and ${\textstyle I}$.

First use Kirchoff’s current law on the upper middle junction: ${\textstyle J_{1}=J_{2}+I}$. Rearranging gives ${\textstyle J_{2}=J_{1}-I}$.

Now, to determine ${\textstyle J_{1}}$, we use Kirchoff’s voltage law on the left loop: ${\textstyle -V_{1}+J_{1}\cdot 1-V_{2}=0}$. Rearranging this and substituting it back into the equation for ${\textstyle J_{2}}$ gives ${\textstyle J_{2}=V_{1}+V_{2}-I}$.

Thus, the second differential equation is

$${\frac {dV_{2}}{dt}}=-{\frac {1}{3}}(V_{1}+V_{2}-I).$$