The steady-state current through $L_1$ after the switch has been closed for a very long time is

After a very long time $\left(\mathrm{t}\to \mathrm{\infty }\right)$, 

$I_0=\frac{V}{R}$          ....(i)

Since the inductors are in parallel,

$L_{\text{1}}\frac{\text{d}{\mathit{\text{I}}}_{\text{1}}}{\text{d}\mathit{\text{t}}}=L_{\text{2}}\frac{\text{d}{\mathit{\text{I}}}_{\text{2}}}{\text{d}\mathit{\text{t}}}$

After a long time,

$L_{\text{1}}{I}_{\text{1}}=L_{\text{2}}{I}_{\text{2}}$         .... (ii)

The total current $I_{\text{0}}$ through the battery is divided into the inductors, hence

$I_0=I_1+I_2$          ..... (iii)

From equations (ii) & (iii)

$I_{\text{1}}=\left(\frac{L_{\text{2}}}{L_{\text{1}}\text{+}{\mathit{\text{L}}}_{\text{2}}}\right)I_{\text{0}}\&I_{\text{2}}=\left(\frac{L_{\text{1}}}{L_{\text{1}}\text{+}{\mathit{\text{L}}}_{\text{2}}}\right)I_{\text{0}}$

The current through $L_{\text{1}}$ is

${\mathit{\text{I}}}_{\text{1}}=\left(\frac{L_{\text{2}}}{L_{\text{1}}+L_{\text{2}}}\right)I_{\text{0}}$

$=\left(\frac{L_{\text{2}}}{L_{\text{1}}+L_{\text{2}}}\right)\frac{V_{\text{0}}}{R}$