Inverting OpAmp with reference voltage

Note that you have an ideal op-amp with zero current into the inputs and with zero voltage between the inputs. Assuming a current \$i(t)\$ through the feedback capacitor (from right to left), we get

$$i(t)=C\frac{dv_o(t)}{dt}\quad\text{and}\quad v_o(t)=\frac{1}{C}\int i(t)dt\tag{1}$$

Since no current flows in or out of the inverting input of the op-amp, the same current also flows through the series connection of \$R\$ and \$C\$ at the input:

$$v_i(t)+Ri(t)+\frac{1}{C}\int i(t)dt=0\tag{2}$$

Plugging \$(1)\$ into \$(2)\$ we get

$$v_i(t)+RC\frac{dv_o(t)}{dt}+v_o(t)=0\tag{3}$$

Using the time constant \$\tau=RC\$, \$(3)\$ can be rewritten as

$$\frac{dv_o(t)}{dt}+\frac{1}{\tau}v_o(t)=-\frac{1}{\tau}v_i(t)\tag{4}$$

The homogeneous solution of \$(4)\$ is

$$v_{o,h}(t)=Ae^{-t/\tau},\quad t>0\tag{5}$$

(with some constant \$A\$ to be determined) and, since \$v_i(t)\$ is constant for \$t>0\$ (equal to \$v_i(0^+)=10V\$), a particular solution is

$$v_{o,p}(t)=-v_i(0^+),\quad t>0\tag{6}$$

The total solution is the sum of the homogeneous and the particular solution:

$$v_o(t)=Ae^{-t/\tau}-v_i(0^+),\quad t>0\tag{7}$$

The constant \$A\$ must be determined from the initial condition \$v_o(0^+)=0\$:

$$v_o(0^+)=A-v_i(0^+)=0\quad\Longrightarrow\quad A=v_i(0^+)\tag{8}$$

which finally gives us the complete solution

$$v_o(t)=v_i(0^+)(e^{-t/\tau}-1),\quad t>0\tag{9}$$

To find the lower cutoff frequency and passband gain, you can use Miller's theorem to separate the \$100nF\$ into two capacitances to ground.

^{simulate this circuit – Schematic created using CircuitLab}

The schematic contains a simulation that shows you the similarity.

I determined \$A_{DC} \approx -16\ 443\$, resulting in an input capacitance of \$C_{in} = C\cdot (1-A_{DC}) \approx 1.6mF\$ and an output capacitance of \$C_{out} = C\cdot \left(1 - \frac{1}{A_{DC}}\right) \approx 100nF\$.

Doing so will allow you to calculate the lower cutoff frequency and passband gain.

$$A_{in\to B,miller} = \frac{R_8C_6s}{1+R_8(C_6+C_9)s}$$

There is a zero at f=0Hz, and a first cutoff frequency at

$$f_{p1} = \frac{1}{2\pi R_8(C_6+C_9)} \approx 0.1Hz$$

The passband gain is severely reduced by the capacitive divider circuit.

$$A_{pb} = A_{DC}\cdot\frac{C_6}{C_6+C_9} \approx 5.75 = 15.2dB$$

At high frequencies, the opamp can't produce high gains and the approximation breaks down. The higher cutoff frequency will be determined by the GBW of the opamp. If the opamp is considered ideal, then this is a high-pass filter.