How to construct a differential equation from this RLC circuit?
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 &ndash; Schematic created using CircuitLab

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

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

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

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

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

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

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

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

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