While solving the simple RC circuit below in a particular way, I came across an interesting mathematical quirk.
simulate this circuit – Schematic created using CircuitLab
Using nodal analysis, we can readily get the expected solution \$v = V_0e^{\frac{-2t}{RC}}\$, where \$V_0\$ is the initial condition.
Since the resistors are in parallel, they can be replaced by a single resistor valued \$R/2\$ and, by KVL, we can arrive at the same result uneventfully. However, it gets at little more interesting if we keep both loops and try to use mesh analysis to solve for one of the currents.
The right loop equation is $$ \frac{1}{C}\int{i_1+i_2 \mathrm{d}t} + Ri_2 = 0\\ RC\frac{\mathrm{d}i_2}{\mathrm{d}t} + i_2 = -i_1 $$ and the left loop equation is, similarly $$ RC\frac{\mathrm{d}i_1}{\mathrm{d}t} + i_1 = -i_2. $$
Solving this system for \$i_2\$, for instance, gives the second-order ODE $$ \frac{\mathrm{d}^2i_2}{\mathrm{d}t^2} + \frac{2}{RC}\frac{\mathrm{d}i_2}{\mathrm{d}t} = 0, $$ which has the solution $$ i_2 = A + Be^{\frac{-2t}{RC}}. $$ Appropriate initial conditions \$i_2(0)\$ and \$\mathrm{d}i_2/\mathrm{d}t(0)\$ can be found, giving \$A=0\$ and \$B=V_0/R\$ as expected. Alternatively, since \$i_2(\infty)\$ has to be 0, we can deduce that \$A=0\$ and only one initial condition is needed to find \$B\$.
While in this case this "phantom" order can be easily avoided, it can be a pitfall for young players and may not be so obvious in more complex circuits. It can also make solving a simpler circuit more difficult. That being said, is there any strategy or best practice that can be used to avoid this apparent extra order creeping up in your analysis?
