Resistor and inductor circuit with a steady voltage source

Question:
A circuit that has a resistor, $R$, and an inductor, $L$, which are connected in series with a DC voltage source $V$. The initial value of the current is $I(t=0)=0$ A. Find the current as a function of time.

Answer:
From Kirchoff's law, the generated voltage is consumed by each element.
\[
V-RI-L\frac{dI}{dt}=0
\]
\begin{eqnarray}
RI+L\frac{dI}{dt} &=& V  \\
I' + \frac{1}{\tau}I &=& \frac{V}{L}
\end{eqnarray}
where $I'=\frac{dI}{dt}$ and $\tau=\frac{L}{R}$. Now, let $D\equiv \frac{d}{dt}$.
\begin{equation}
\left(D+\frac{1}{\tau}\right)I=\frac{V}{L}  \\
\end{equation}
The fundamental solution is
\begin{equation}
I_f = C_1e^{-\frac{t}{\tau}}
\end{equation}
The particular solution is
\begin{equation}
I_p = \frac{1}{D+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{1}{0+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{V}{R}
\end{equation}
Therefore, we have the general solution.
\begin{equation}
I = C_1e^{-\frac{t}{\tau}} + \frac{V}{R}
\end{equation}
When $t=0$, $I=0$. So $0=C_1+V/R$ and $C_1=-V/R$. Then, the current is expressed as
\[
I(t) = -\frac{V}{R}e^{-\frac{t}{\tau}}+\frac{V}{R}
\]
or
\[
I(t) = \frac{V}{R}(1-e^{-\frac{t}{\tau}})
\]

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