Problem 9.4-3

Let’s separate the current into i_S (for the source, on the left) and i_C (for the capacitor, on the right). We have i_L = i_C + i_S

We have, for the KVL in the left loop:

Vs = 1*i_S + 2* i_L + L*di_L/dt

But i_L = i_C + i_S so Vs = i_S + 2*(i_C + i_S)+ L*di_L/dt

And replacing i_S = i_L – i_C

We get Vs = 3*i_L – i_C + L*di_L/dt  (1)

For the right side loop, we have:

V_C = 2*i_L + L*di_L/dt and we have i_C = - C*dV_C/dt

Thus i_C = - C*(2di_L/dt+ L*d²i_L/dt²)

By replacing this i_C in the equation (1) we get:

Vs = 3*i_L + C*(2di_L/dt+ L*d²i_L/dt²)+ L*di_L/dt

LC*d²i_L/dt2 + (L + 2C)*di_L/dt + 3*i_L = Vs

The roots are given by r± solution of this equation,