The constant particular solution is $A=V/(4R)$, and the homogeneous equation has decay rate $2R/L$. Imposing $I_1(0)=0$ gives
$$
\boxed{I_1(t)=\frac{V}{4R}\left(1-e^{-t/\tau}\right),\qquad A=\frac V{4R},\quad\tau=\frac L{2R}.}
$$
As a separate check, the <Thévenin equivalent> seen by $L$ has open-circuit voltage $V/2$ and resistance $R+(2R\parallel2R)=2R$, again giving $\tau=L/(2R)$.

Solution

Past exam of the physics 2 paper of the University of Oxford / 2014 / 9 / ii / Solution

 Solution (source code)