= Solution
The source diagram has $2R$ in series with two parallel branches: one containing $R$ then $L$, and the other containing $2R$. Let their common voltage be $u(t)$, measured from the upper node to the lower return. <Kirchhoff's current law> and the branch laws give
$$
I_0=I_1+I_2,\qquad u=2RI_2=RI_1+L\frac{dI_1}{dt},\qquad V=2RI_0+u.
$$
Using $I_2=u/(2R)$ in the source equation yields $u=V/2-RI_1$. Substitution into the coil branch gives
$$
\boxed{L\frac{dI_1}{dt}+2RI_1=\frac V2.}
$$
The switch was open for a long time, so $I_1(0^-)=0$. <Inductor current continuity> gives $I_1(0^+)=0$ for a finite applied voltage.
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