Node Voltage Problem, possible to simplify?
I think you need \$-3\cdot V_1\$ in the first column, third row; removing it from the third column, third row. (It's incoming, so negative, if I take your other signs to be correct.) Since the first column implies \$V_1\$, you'd just put \$-3\$ there, not \$-3\cdot V_1\$, in keeping with your notation. I think you need to associate the columns with the nodes and you didn't do that, there. The rest of that row looks good to me.

I looked at the first and second rows and they look good to me, as well. I didn't look over the fourth row, but since you didn't ask about it and you've done really well elsewhere, I'll leave it unchecked for now.

The second row I got was:

$$\begin{align*}
\left[-\frac{V_1}{R_2}\right] + \left[\frac{V_2}{R_2} + C_1\frac{\textrm{d}V_2}{\textrm{d}t} + C_2\frac{\textrm{d}V_2}{\textrm{d}t}\right] +\left[- C_2\frac{\textrm{d}V_3}{\textrm{d}t}\right] +\left[0\cdot V_4\right]&amp;= 0
\end{align*}$$

The third row I got was:

$$\begin{align*}
\left[-3\cdot V_1\right] +\left[- C_2\frac{\textrm{d}V_2}{\textrm{d}t}\right] + \left[\frac{V_3}{R_3} + C_2\frac{\textrm{d}V_3}{\textrm{d}t} + \frac{1}{L_1}\int V_3\; \textrm{d}t\right] - \left[\frac{1}{L_1}\int V_4\; \textrm{d}t\right] = 0
\end{align*}$$