RL circuit Step-Response Equation explaination
The solution approach is just a bog standard 1st order diff eq solution. Probably in chapter 1 of any diff eq book. But it sounds more like you don't see how to integrate an exponential. Which is even earlier -- like MTH252, I think.

The standard form is just as given:

$$\frac{\textrm{d}y}{\textrm{d}t} + P_x\cdot y = Q_x$$

If you can set things up like that, then your integrating factor (which is a nifty way to solve these) is:

$$\mu=e^{\int P_x\; \textrm{d}x}$$

Then then the solution is:

$$ y=\frac{1}{\mu}\int \mu\cdot Q_x\;\; \textrm{d}x$$

You can see the minor trouble that they went to in order to put it into standard form, through they called one of the terms \$r_x\$ instead. Doesn't matter. So let's just look at the solution:

$$\begin{align*}
i_t &amp;= \frac{1}{\mu}\int \mu\cdot Q_x\;\; \textrm{d}t,~~~~~\mu= e^{\int P_x \textrm{d}x}=e^{\int \frac{R}{L} \textrm{d}x}=e^{\frac{R}{L}x}\\
i_t &amp;= \frac{1}{e^{\frac{R}{L}t}}\int e^{\frac{R}{L}x}\cdot \frac{V_s}{L}\;\; \textrm{d}x \\
i_t &amp;= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{L}\int e^{\frac{R}{L}x} \;\; \textrm{d}x \\
i_t &amp;= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{L}\left[\frac{L}{R}\left(e^{\frac{R}{L}x}\right)\right]\Biggr\vert_0^{t} \\
i_t &amp;= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{R}\left[e^{\frac{R}{L}x}\right]\biggr\vert_0^{t} \\
i_t &amp;= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{R}\left[e^{\frac{R}{L}t} - 1\right] \\
i_t &amp;= \frac{V_s}{R}\left[1 - \left(e^{-\frac{R}{L}t}\right)\right] =\frac{V_s}{R} - \frac{V_s}{R}\left(e^{-\frac{R}{L}t}\right)~~~~~~\tau=\frac{L}{R} \\
\therefore i_t &amp;= \frac{V_s}{R} - \frac{V_s}{R}\left(e^{-\frac{t}{\tau}}\right)
\end{align*}$$

I hope that helps.