In the circuit shown below, i _ 1 and i _ 2 are the steady-state...

Since the potential difference across the inductors is same, we get

${L}_{1}\dfrac{d{i}_{1}}{dt} = {L}_{2}\dfrac{d{i}_{2}}{dt}$

${L}_{1}d{i}_{1} = {L}_{2}d{i}_{2}$

${L}_{1} ∆ {i}_{1} = {L}_{2} ∆ {i}_{2}$

${L}_{1}{i}_{1} = {L}_{2}{i}_{2}$ (Since the initial value of current through both the inductors was zero)&

$\dfrac{i1}{i2} = \dfrac{L2}{L1}$

The steady current $i$ passing through $R$ is

$\text{i} = \dfrac{E}{R}$

${i}_{1} = \left(\dfrac{L2}{{L}_{1} + {L}_{2}}\right)i = \dfrac{E{L}_{2}}{R\left({L}_{1} + {L}_{2}\right)}$