It seems to me you exchanged inverting and non-inverting inputs of both op-amps. Supposing to reverse the polarity we have:
inverting input 1: V(1-)=Vo1/2
non-inverting input 1: V(1+)=V1
inverting input 2: V(2-)=Vo2+(Vo1-Vo2)*(1/SC)/[(1/SC)+R] = Vo2+(Vo1-Vo2)/(SCR+1)
non-inverting input 2: V(2+)=V1
From the first two equations: Vo1=2V1
From the last two: Vo2*[1-1/(SCR+1)]+Vo1/(SCR+1)=V1, since Vo1=2V1 we have:
Vo2*[1-1/(SCR+1)]+2V1/(SCR+1)=V1 then:
Vo2=[1-2/(SCR+1)]/[1-1/(SCR+1)]*V1 simplifying:
Vo2=V1*(SCR-1)/SCR
From the schematic I=(V1-Vo2)/R then:
I=V1*[1-(SCR-1)/SCR]/R that is:
I=V1/SCR² Then:
V1/I=SCR²
Since the laplace equation of an inductor is V/I=SL then the circuit will simulate an inductor of value L=CR².
Please check the math is correct.