As usual, we call V1 the inverting input node voltage and V2
the non-inverting input node voltage.
The two currents entering the op amp are equal to zero.
We have Vs = 15*u(t)*exp(-2t) and Vs + R*I(t) = V1
But the op amp tells us that V1 = V2 = 0 thus we
have Vs(t) = - 15.103*I(t)
Thus we deduce that I(t) = u(t)*exp(-2t) in mA
Let’s call Ic(t) the current in the upper branch (in the capacitor) and
Ir(t) the current in the lower branch (in the resistor).
We have I(t) = Ic(t) + Ir(t)
We also have Vo(t) = - 103*Ir(t) = - 103*[I(t)
– Ic(t)]
But Ic(t) = - C*dVo(t)/dt so Vo(t) = - 103*[C*dVo(t)/dt
+ I(t)]
C = 0.25.10-6 F
So we get a differential equation:
0.25.10-3*dVo(t)/d + Vo(t) = I(t) = 10-3*u(t)*exp(-2t)
Which is also dVo(t)/d + 4.103*Vo(t) = u(t)*exp(-2t)
The solution to this equation is Vo(t) = u(t)*exp(-2t) + A*exp(-4.103*t)
For A, we need Vo(0) = A = 10 V
Thus the solution is Vo(t) = u(t)*exp(-2t) + 10*exp(-4.103*t)