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): $$i_3=\frac{1}{R_2C_1}\int{i_2dt}+\left(\frac{R_1}{R_2}+1\right)i_2-\frac{L}{R_2}\frac{d(i_1-i_2)}{dt}$$ $$i_3=\frac{L}{R_3}\frac{d(i_1-i_2)}{dt}+i_1-\frac{v_1}{R_3}$$ $$i_3=\frac{1}{R_2+R_3}\left(\frac{1}{C_1}\int{i_2dt}+(R_1+R_2)i_2+i_1R_3-v_1\right)$$ $$\frac{di_1}{dt}=\frac{di_2}{dt}+\frac{R_1}{L}i_2+\frac{1}{LC_1}\int{i_2dt}+\frac{R_2}{L}(i_2-i_3)$$ $$i_1=\left(\frac{R_2+R_4}{R_3}+1\right)i_3+\frac{1}{R_3C_2}\int{i_3dt}-\frac{R_2}{R_3}i_2$$ $$i_1=\frac{1}{R_3}\left(v_1-i_2R_1-\frac{1}{C_1}\int{i_2dt}-i_2R_2+(R_2+R_3)i_3\right)$$ $$\frac{di_1}{dt}=\frac{v_1}{L}+\frac{di_2}{dt}+\frac{R_2}{L}i_2-\frac{R_2+R_4}{L}i_3-\frac{1}{LC_2}\int{i_3dt}$$

  • 5
    \$\begingroup\$ Use Laplace transform. It makes your integro-differential equations simple linear equations. \$\endgroup\$
    – nidhin
    Commented Apr 16, 2018 at 11:49
  • \$\begingroup\$ I'm looking for how to derive the differential equation, not to solve it, in which the Laplace transform can be applied as you have mentioned. The problem is, I cannot find a differential equation which solves only for \$i_2\$ (I always end up with differential equations with \$i_1\$ or \$i_3\$ or both in it which I can't get rid of). \$\endgroup\$
    – mjtsquared
    Commented Apr 16, 2018 at 12:57
  • 3
    \$\begingroup\$ I think you can derive an equation for \$i_2\$ in Laplace domain and then apply inverse transform to get the differential equation. \$\endgroup\$
    – nidhin
    Commented Apr 16, 2018 at 13:23

1 Answer 1


I really think nidhin's answer is the best way to go. Use the Laplace transform, calculate \$I_2(s)\$ and then convert it back into a differential equation if you're not interested in the solution.

I just did it for you to show you what you're up against if you try to avoid Laplace transforms:

The Laplace transform gives the solution (I used a CAS for this):

\$I_2\cdot (a\cdot s^3 + b\cdot s^2 + c\cdot s + d) = V_1\cdot (e\cdot s^3 + f\cdot s^2)\$

Yielding the differential equation of the form:

\$a\frac{d^3i_2}{dt^3} + b\frac{d^2i_2}{dt^2} + c\frac{di_2}{dt} + d\cdot i_2 = e\frac{d^3v_1}{dt^3} + f\frac{d^2v_1}{dt^2}\$

Where the coefficients are (I hope I copied them correctly):

\$a = C_1C_2L(R_3R_4 + R_2R_4 + R_1R_4 + R_1R_2)\$

\$b = C_1C_2R_3(R_2R_4 + R_1R_4 + R_1R_2) + (C_1+C_2)L(R_2 + R_3) + L(C_2R_4 + C_1R_1)\$

\$c = C_2R_3(R_4 + R_2) + C_1R_3(R_2 + R_1)\$

\$d = R_3\$

\$e = C_1C_2L(R_2+R_3+R_4)\$

\$f = C_1(C_2R_2R_3+L)\$

I would not advise doing this without transforming to Laplace. While it is possible to solve it without, it can be very hard to see how the equations need to be substituted.


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