What is the IL(t) of the RL Circuit of the IL(t=0) = I ini
I wish you'd exposed more details of how you walked through your process. But at least you show some work. Rather than point to an error, I'd rather simply walk through the process as I'd see it, instead.

Schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

Nodal analysis provides:

$$\begin{align*}
\frac{V_X}{R_1}+\frac{1}{L_1}\int V_X\:\text{d}t&amp;= \frac{V_S}{R_1}
\end{align*}$$

Taking the derivative, I get:

$$\begin{align*}
\frac{\text{d} V_X}{R_1\:\text{d}t}+\frac{V_X}{L_1}&amp;= 0\\\\
\frac{\text{d} V_X}{\text{d}t}&amp;= -\frac{R_1\:V_X}{L_1}\\\\
\frac{\text{d} V_X}{V_X}&amp;= -\frac{R_1}{L_1}\:\text{d}t\\\\
\int \frac{\text{d} V_X}{V_X}&amp;= \int-\frac{R_1}{L_1}\:\text{d}t\\\\
\operatorname{ln}\left(V_X\right)&amp;= -\frac{R_1}{L_1}\:t+C_0\\\\
V_X &amp;= A_0\:e^{-\frac{R_1}{L_1}\:t}
\end{align*}$$

The initial conditions are that \$V_{X_{t=0}}=A_0=V_S-I_0\cdot R_1\$. So, the above becomes:

$$\begin{align*}
V_X &amp;= \left(V_S-I_0\cdot R_1\right)\:e^{-\frac{R_1}{L_1}\:t}
\end{align*}$$

You know that the current must always be \$I_X=\frac{V_S-V_X}{R_1}\$, so:

$$\begin{align*}
I_X &amp;= \frac{V_S-\left(V_S-I_0\cdot R_1\right)\:e^{-\frac{R_1}{L_1}\:t}}{R_1}\\\\
&amp;=\frac{V_S}{R_1}\cdot\left(1-e^{-\frac{R_1}{L_1}\:t}\right)+I_0\cdot e^{-\frac{R_1}{L_1}\:t}
\end{align*}$$

It is handy to set \$\tau=\frac{L_1}{R_1}\$. Then:

$$\begin{align*}
I_X&amp;=\frac{V_S}{R_1}\cdot\left(1-e^{-\frac{t}{\tau}}\right)+I_0\cdot e^{-\frac{t}{\tau}}
\end{align*}$$

If you figure \$\tau=1\:\mu\text{s}\$, \$I_0=3\:\text{mA}\$, \$V_S=10\:\text{V}\$, \$R_1=1\:\text{k}\Omega\$, and plug in \$t=0\$ you will find \$I_{X_{t=0}}=3\:\text{mA}\$ (Also, \$I_{X_{t=1\:\mu\text{s}}}\approx 7.4\:\text{mA}\$ and \$I_{X_{t=2\:\mu\text{s}}}\approx 9.1\:\text{mA}\$, etc.)