So I was drawing arbitrary RLC networks and solving them as an exercise. But I came across this one and it seems impossible.
Specifically, I want to derive a differential equation that solves for \$i_2\$ (an equation with \$i_2\$ and its derivatives).
simulate this circuit – Schematic created using CircuitLab
And here's my attempt. The voltages and currents are defined such that: $$v_2=L\frac{d(i_1-i_2)}{dt}, v_3=i_2R_1, v_4=\frac{1}{C_1}\int{i_2dt}, v_5=(i_2-i_3)R_2,$$ $$v_6=(i_1-i_3)R_3, v_7=i_3R_4, v_8=\frac{1}{C_2}\int{i_3dt}$$ Applying Kirchhoff's voltage law to all loops gives us: $$v_2=v_3+v_4+v_5=L\frac{d(i_1-i_2)}{dt}=i_2R_1+\frac{1}{C_1}\int{i_2dt}+(i_2-i_3)R_2$$ $$v_6=v_7+v_8-v_5=(i_1-i_3)R_3=i_3R_4+\frac{1}{C_2}\int{i_3dt}-(i_2-i_3)R_2$$ $$v_1=v_2+v_6=L\frac{d(i_1-i_2)}{dt}+(i_1-i_3)R_3$$ $$v_1=v_3+v_4+v_5+v_6=i_2R_1+\frac{1}{C_1}\int{i_2dt}+(i_2-i_3)R_2+(i_1-i_3)R_3$$ $$v_1=v_2-v_5+v_7+v_8=L\frac{d(i_1-i_2)}{dt}-(i_2-i_3)R_2+i_3R_4+\frac{1}{C_2}\int{i_3dt}$$ $$v_1=v_3+v_4+v_7+v_8=i_2R_1+\frac{1}{C_1}\int{i_2dt}+i_3R_4+\frac{1}{C_2}\int{i_3dt}$$ $$v_2+v_6=v_3+v_4+v_7+v_8$$ $$=L\frac{d(i_1-i_2)}{dt}+(i_1-i_3)R_3=i_2R_1+\frac{1}{C_1}\int{i_2dt}+i_3R_4+\frac{1}{C_2}\int{i_3dt}$$ Now all I need to do is somehow write \$i_1\$ and \$i_3\$ in terms of \$i_2\$, and if the above clues aren't enough to do just that, I don't know what else. But it seemingly cannot be done!..as far as I have tried! Here are a few definitions of \$i_3\$ and \$i_1\$ that I've picked up using above clues (which you can hopefully build up on): $$i_3=\frac{1}{R_2C_1}\int{i_2dt}+\left(\frac{R_1}{R_2}+1\right)i_2-\frac{L}{R_2}\frac{d(i_1-i_2)}{dt}$$ $$i_3=\frac{L}{R_3}\frac{d(i_1-i_2)}{dt}+i_1-\frac{v_1}{R_3}$$ $$i_3=\frac{1}{R_2+R_3}\left(\frac{1}{C_1}\int{i_2dt}+(R_1+R_2)i_2+i_1R_3-v_1\right)$$ $$\frac{di_1}{dt}=\frac{di_2}{dt}+\frac{R_1}{L}i_2+\frac{1}{LC_1}\int{i_2dt}+\frac{R_2}{L}(i_2-i_3)$$ $$i_1=\left(\frac{R_2+R_4}{R_3}+1\right)i_3+\frac{1}{R_3C_2}\int{i_3dt}-\frac{R_2}{R_3}i_2$$ $$i_1=\frac{1}{R_3}\left(v_1-i_2R_1-\frac{1}{C_1}\int{i_2dt}-i_2R_2+(R_2+R_3)i_3\right)$$ $$\frac{di_1}{dt}=\frac{v_1}{L}+\frac{di_2}{dt}+\frac{R_2}{L}i_2-\frac{R_2+R_4}{L}i_3-\frac{1}{LC_2}\int{i_3dt}$$