Differential equation from a Wheatstone bridge with Inductor
Hints: 

Voltage at left voltage divider (between R1 and R2) is just input voltage scaled down by a constant factor \$\frac{R_2}{R_1+R_2}\$. 

Voltage at right terminal (between R3 and R4) is R4 times current through inductor \$I_L\$. Get current through inductor by setting up an appropriate diffenrential equation using \$U_L=-L \frac{dI_L}{dt}\$ and KVL and Ohm's law.