Inductor time variable current
The general nodal equation is:
$$\frac{V_t}{R_1}+\frac1{L_1}\int V_t\:\text{d}t+I_{L\left(t=t_i\right)}=I_{t=t_i}\tag{1}\label{e1}$$
Where \$I_{t=t_i}\$ is your current source value at \$t=t_i\$ and \$I_{L\left(t=t_i\right)}\$ is the inductor current's initial condition at \$t=t_i\$. I'm keeping this in a general form, since you'll need it twice: at \$t=t_{_0}=0^+\,\text{s}\$ and at \$t=t_{_1}\$.
Putting this into differential form gives:
$$\frac{\text{d}}{\text{d}\, t}V_t+ \frac{R_1}{L_1}\,V_t=0\:\text{V}\cdot\text{s}^{-1}\tag{2}\label{e2}$$
For your first situation at \$t_{_0}\$, you know that the initial condition is \$V_{t=0^+\,\text{s}}=I_1\,R_1\$. So you can solve the differential equation for \$V_t\:\mid_0^{t_1}\$. And it immediately follows from KCL that \$I_{L\left(t\right)}\mid_0^{t_1}=I_1-\frac{V_t\:\mid_0^{t_1}}{R_1}\$, from which you can develop the equation for the current in the inductor.
Similarly, the exact same equation found in Eq. \$\ref{e2}\$ also still applies in your newer situation. However, for \$t\ge t_{_1}\$, you will have a new initial condition, \$V_{t_{_1}}\:\mid_0^{t_1}\$ (which you can easily compute from the earlier voltage equation.) Getting a result here is the same process, though. Just don't forget to do your substitutions to simplify and keep track of your time offsets. Keep in mind this is a new \$V_{t}\:\mid_{t_1}^{\infty}\$ equation. Again, it immediately follows from KCL that \$I_{L\left(t\right)}\mid_{t_1}^{\infty}=I_2-\frac{V_{t}\:\mid_{t_1}^{\infty}}{R_1}\$.