Positive Feedback in Time Domain
There are two possible states. In both cases, \$V_2\$ is irrelevant since it is finite.


If \$V_O=+\infty\$ then \$V_2\$, being finite, can be ignored. \$V_+=\frac{+\infty}{2}\$ and \$V_-=\frac{+\infty}{3}\$. Since \$V_+\gt V_-\$, the opamp holds \$V_O=+\infty\$.
If \$V_O=-\infty\$ then \$V_2\$, being finite, can be ignored. \$V_+=\frac{-\infty}{2}\$ and \$V_-=\frac{-\infty}{3}\$. Since \$V_+\lt V_-\$, the opamp holds \$V_O=-\infty\$.


The above follows because:

$$\begin{align*}
V_+&amp;=\frac{V_O}{2}+V_2\\\\
V_-&amp;=\frac{V_O}{3}
\end{align*}$$

It's a binary state that depends on the initial condition of the output.



If you limit the output to a pair of voltage rails you provide, \$V_\text{CC}\$ and \$V_\text{EE}\$, then the situation is different. You have a range of hysteresis. Starting \$V_2\$ from well below \$\frac{-V_\text{CC}}{6}\$, and raising it upwards, the output will transition from \$V_\text{EE}\$ to \$V_\text{CC}\$ at \$V_2=\frac{-V_\text{EE}}{6}\$. Starting \$V_2\$ from well above \$\frac{-V_\text{EE}}{6}\$, and lowering it downwards, the output will transition from \$V_\text{CC}\$ to \$V_\text{EE}\$ at \$V_2=\frac{-V_\text{CC}}{6}\$.

In the somewhat more realistic case where also the open loop gain is large, but finite, then there will be a very tiny linear transition slope nearby the above-mentioned transition points. (And assuming an even more realistic opamp with BJT inputs, there will be substantial input loading for the input nodes when the voltages at the nodes are more than a few tens of millivolts apart from each other, since the differential amplifier inputs only present a tiny load in the case where the inputs are very close to each other in voltage.)