Op-amp circuit analysis with capacitors on feedback and input lines
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

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

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

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

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

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

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

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