Is my approach for the RL circuit correct?
The answer is \$I_{L\left(t\right)}=10\:\text{A}\cdot\left(1-e^{\frac{R}{L}t}\right)\$.

It follows from:

$$\begin{align*}
\frac{V_t}{R}+\frac{1}{L}\int V_t\:\text{d}t&amp;=I_{s\left(t\right)}\\\\
\frac{1}{R}\frac{\text{d} V_t}{\text{d}t}+\frac{V_t}{L}&amp;=\frac{\text{d}I_{s\left(t\right)}}{\text{d} t}=0
\end{align*}$$

During the period from \$t=0\$ to \$t=100\:\mu\text{s}\$, \$\frac{\text{d}I_{s\left(t\right)}}{\text{d} t}=0\$.

So, it follows for this period that:

$$\begin{align*}
\frac{\text{d} V_t}{\text{d}t}&amp;=-\frac{R}{L}V_t\\\\
\frac{\text{d} V_t}{V_t}&amp;=-\frac{R}{L}\text{d}t\\\\
\int\frac{\text{d} V_t}{V_t}&amp;=-\frac{R}{L}\int\text{d}t\\\\
\operatorname{ln}\left(V_t\right)&amp;=-\frac{R}{L}t + C\\\\
V_t&amp;=e^C\cdot e^{-\frac{R}{L}t}
\end{align*}$$

Given the initial conditions at \$t=0\$, then:

$$V_t=R\cdot 10\:\text{A}\cdot e^{-\frac{R}{L}t}$$

Now, it's easy. We know that;

$$\begin{align*}
I_{L\left(t\right)}&amp;=\frac{1}{L}\int V_t\:\text{d} t\\\\
&amp;=\frac{1}{L}\int R\cdot 10\:\text{A}\cdot e^{-\frac{R}{L}t}\:\text{d} t\\\\
&amp;=10\:\text{A}\cdot\frac{R}{L}\int e^{-\frac{R}{L}t}\:\text{d} t\\\\\
&amp;=10\:\text{A}\cdot\frac{R}{L}\frac{L}{R}\left(1- e^{-\frac{R}{L}t}\right)+C_0\\\\
&amp;=10\:\text{A}\cdot\left(1- e^{-\frac{R}{L}t}\right)+C_0\\\\
\end{align*}$$

Here, \$C_0=0\$, so:

$$\begin{align*}
I_{L\left(t\right)}&amp;=10\:\text{A}\cdot\left(1- e^{-\frac{R}{L}t}\right)\\\\
\end{align*}$$

So, at \$t=100\:\mu\text{S}\$, then \$I_{L\left(t=100\:\mu\text{s}\right)}= 6.3212\:\text{A}\$

Also,

$$V_{t=100\:\mu\text{s}}=R\cdot 10\:\text{A}\cdot e^{-\frac{R}{L}t}= 183.94\:\text{V}$$

You can work out the rest after \$t \gt 100\mu\text{s}\$.