question with rl circuit
When the switch is opened, inductor acts like a current source with \$I_o\$. Where \$I_o \$ is the current which was flowing through resistor 15 kohms, at steady state ( when switch is closed for long time, inductor can be considered a short).
Then you can use RL circuit current equation to solve for current at any time t.
$$I_1(t)=I_2(t)=I(t) = I_oe^{-tR/L}$$ where R = 15k + 5k ohms