How to construct a differential equation from this RLC circuit?

Question

I'm trying to figure out how to construct a differential equation for the natural response of the following circuit and I am having trouble. The capacitor is initially charged with a voltage \$v_0\$ and I want to solve for \$v(t)\$, the voltage across the capacitor.

^{simulate this circuit – Schematic created using CircuitLab}

Here is my attempt so far:

By Kirchhoff's laws, \$v_L=v_2\$, \$v=v_1+v_L\$, and \$i_1+i_2+i_3=0\$. The currents and voltages are defined so that \$i_1=-\frac{v_1}{R_1}=C\frac{dv}{dt}\$, \$i_2=\frac{1}{L}\int{v_Ldt}\$, and \$i_3=\frac{v_2}{R_2}=\frac{v_L}{R_2}\$. $$C\frac{dv}{dt}+\frac{1}{L}\int{v_Ldt}+\frac{v_L}{R_2}=0$$ And I cannot come up with an equation the exclusively solves for \$v\$. Is there even an existing solution to this problem?

Answer 1

You just need to add one more equation

$$v_1 = -i_1 R_1$$ $$v_1 = -CR_1\frac{dv}{dt}$$

Thus $$ v_L = v - v_1 = v + C R_1 \frac{dv}{dt}$$

Now you can substitute in your integro-differential equation to get an equation with just one variable

$$C\frac{dv}{dt} + \frac{1}{L}\int_{-\infty}^t (v + CR_1\frac{dv}{dt})dt +\frac{v+CR_1\frac{dv}{dt}}{R_2} = 0$$

From here you can rearrange terms and take derivatives until you reach a differential equation in one variable, that can be solved by the usual methods (recognizing a known form, Laplace transform, etc.)

Answer 2

Just so you have my approach to consider, as well. (I know you've already selected an answer.) Here's the redrawn schematic which I prefer:

^{simulate this circuit – Schematic created using CircuitLab}

I apply nodal analysis and get these two equations from the two nodes:

$$\begin{align*} \frac{V_\text{C}}{R_1}+C\frac{\text{d} V_\text{C}}{\text{d} t}&= \frac{V_\text{L}}{R_1}\label{n1}\tag{node $V_\text{C}$}\\\\ \frac{V_\text{L}}{R_1}+\frac{V_\text{L}}{R_2}+\frac{1}{L}\int V_\text{L}\:\text{d} t&=\frac{V_\text{C}}{R_1}\label{n2}\tag{node $V_\text{L}$} \end{align*}$$

Just solve the \$\ref{n1}\$ equation for \$V_\text{L}\$:

$$\begin{align*} V_\text{L}&=V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}\label{n3}\tag{solved for $V_\text{L}$} \end{align*}$$

, and then substitute that into the above for \$\ref{n2}\$:

$$\begin{align*} \frac{V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}}{R_1}+\frac{V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}}{R_2}+\frac{1}{L}\int \left[V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}\right]\:\text{d} t&=\frac{V_\text{C}}{R_1} \\\\ \frac{V_\text{C}}{R_1}+\frac{V_\text{C}}{R_2}+C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d} V_\text{C}}{\text{d} t}+\frac{1}{L}\left[\int V_\text{C}\:\text{d} t+R_1\:C\int\text{d} V_\text{C}\right]&=\frac{V_\text{C}}{R_1}\\\\ \frac{V_\text{C}}{R_2}+C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d} V_\text{C}}{\text{d} t}+\frac{1}{L}\left[\int V_\text{C}\:\text{d} t+R_1\:C\int\text{d} V_\text{C}\right]&=0 \end{align*}$$

, now take everything with respect to the derivative of time:

$$\begin{align*} \frac{1}{R_2}\frac{\text{d} V_\text{C}}{\text{d} t}+C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d}^2 V_\text{C}}{\text{d} t^2}+\frac{1}{L}\left[V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}\right]&=0\\\\ C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d}^2 V_\text{C}}{\text{d} t^2}+\left(\frac{1}{R_2}+\frac{R_1\:C}{L}\right)\frac{\text{d} V_\text{C}}{\text{d} t}+\frac{V_\text{C}}{L}&=0\\\\ \end{align*}$$

You can easily put that into standard form and solve using the usual 2nd order diff-eq approach or else use Laplace.