Doing nodal analysis on a circuit with a capacitor.
The left side schematic is at \$t=0^-\$ and the right side schematic is at \$t=0^+\$:







simulate this circuit &ndash; Schematic created using CircuitLab

The above was figured out by mere inspection. I didn't bother with nodal analysis. It was sufficient to see that from either \$V_A\$'s or \$V_B\$'s perspective, \$R_2\$, \$R_3\$, and \$R_4\$ always total out to \$2\:\text{k}\Omega\$ to ground. For the left side, this meant \$V_A=\frac13\cdot36\:\text{V}=12\:\text{V}\$ and the rest just falls out. For the right side,  \$I_{t=0^+}=\frac{12\:\text{V}-6\:\text{V}}{2\:\text{k}\Omega+2\:\text{k}\Omega}=1.5\:\text{mA}\$ so it followed that \$V_B=2\:\text{k}\Omega\cdot 1.5\:\text{mA}=3\:\text{V}\$ and the rest just falls out quickly, again.

If you were doing nodal analysis for \$t=0^-\$ (left side schematic) then:

$$\begin{align*}
\frac{V_A}{R_1}+\frac{V_A}{R_2}+\frac{V_A}{R_3}&amp;=\frac{36\:\text{V}}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{V_B}{R_3}\\\\
\frac{V_B}{R_3}+\frac{V_B}{R_4}&amp;=\frac{V_A}{R_3}+\frac{0\:\text{V}}{R_4}
\end{align*}$$

Which, of course, solves out as \$V_A=12\:\text{V}\$ and \$V_B=6\:\text{V}\$. No shock. Also, since there is no current in \$R_5\$, it follows that \$V_{C_1}=6\:\text{V}\$ at \$t=0^-\$.

If you were doing nodal analysis for \$t=0^+\$ (right side schematic), and temporarily treating \$C_1\$ as a voltage source where \$V_{C_1}=6\:\text{V}\$ at \$t=0^+\$:

$$\begin{align*}
\frac{V_A}{R_2}+\frac{V_A}{R_3}&amp;=\frac{0\:\text{V}}{R_2}+\frac{V_B}{R_3}\\\\
\frac{V_B}{R_3}+\frac{V_B}{R_4}&amp;=I_{C_1}+\frac{V_A}{R_3}+\frac{0\:\text{V}}{R_4}\\\\
\frac{V_C}{R_5}+I_{C_1}&amp;=\frac{12\:\text{V}}{R_5}\\\\
V_C&amp;=V_B+V_{C_1}
\end{align*}$$

Which, of course, solves out as \$V_A=1.5\:\text{V}\$, \$V_B=3\:\text{V}\$, \$V_C=9\:\text{V}\$, and \$I_{C_1}=1.5\:\text{mA}\$. Again, no shock.

All that remains, really, is to develop these equations for \$t\gt 0\$. But that's not part of your question.



From the above, it's easy to show the value of \$i_x\$. And it's just as you say it is.