Digital to Analog Converter math question

Here is some insight. The + input of the op-amp is grounded, and the negative feedback action has the effect of keeping the - input of the op-amp at the same voltage as the + input. So the - input is a virtual ground.

You can solve this by pretending that the op-amp is not there, and then solving the remaining network of resistors and current source by finding the value of \$V_o\$ which is necessary for the junction between the \$3\Omega\$ and \$2\Omega\$ resistors to be at zero volts.

The easiest way to do this is to pretend that the junction is actually grounded (not simply at 0V potential). Determine how much current is dumped into that ground through the \$2\Omega\$ resistor. But then, recognize that there is no ground there and so the current flowing toward that node actually returns via the \$3\Omega\$ feedback resistor.

The current across the feedback resistor gives you a straightforward way to determine the voltage across that resistor, which directly determines \$V_o\$.

I'm not familiar with multisim, but from the look of things, you're driving a set of op-amps powered with 0-3.3V with a 1V P-P sine-wave centered around 0V (which means it swings from +0.5V to -0.5V).

As such, your op-amps will be railed for 1/2 of the input cycle, which will make your filter behave very strangely. I suspect the fact that it looks like it's working at all may be an issue with the simulation not being very accurate, or behaving strangely when used outside of the op-amp's "proper" bounds of operation.

You need to add some DC bias to your input, so it idles at \$ \frac{Vpp-Vss}{2}\$.