RL voltage across inductor differential equation
You got the second relation wrong:

\$ i(t) = \dfrac{v_{in}(t)-v_{out}(t)}{R} \$

substituting in the first equation you get:

\$ v_{in}(t) =
 \dfrac{L}{R}\left(\dfrac{dv_{in}}{dt} - \dfrac{dv_{out}}{dt}\right) + v_{in}(t) - v_{out}(t) \$

Thus \$v_{in}\$ cancels out in both members. Rearranging you get:

\$
\dfrac{L}{R}\dfrac{dv_{out}}{dt} + v_{out} = \dfrac{L}{R} \dfrac{dv_{in}}{dt}
\$

Hence:

\$
\dfrac{L}{R} \cdot j\omega V_{out} + V_{out} = \dfrac{L}{R} \cdot j\omega V_{in}
\$

\$
H(\omega) = \dfrac{V_{out}}{V_{in}} = 
\dfrac{ \dfrac{L}{R} j\omega } { 1 + \dfrac{L}{R} j\omega }
\$

This is coeherent with the standard s-domain approach for determining the transfer function of the system \$W(s)\$. In fact if you replace the inductor with its s-domain equivalent \$sL\$ and apply the voltage divider formula in the s-domain you get:

\$
V_{out}(s) = V_{in}(s) \dfrac{Ls}{R+Ls}
\qquad
\Leftrightarrow
\qquad
W(s) = \dfrac{V_{out}(s)}{V_{in}(s)} = \dfrac{Ls}{R+Ls} 
= \dfrac{\dfrac{L}{R}s}{1+\dfrac{L}{R}s}
\$

and since 

\$
H(\omega)=W(j\omega)=\dfrac{ \dfrac{L}{R} j\omega } { 1 + \dfrac{L}{R} j\omega }
\$

you can see that the two methods give the same result.