This is the work in question.. specifically when I back subsitute $O$ for $y$, I made a mistake in how I handle $k_{1}$, however, it still needs to come out with the form $O = O_{o} \: e^{-t/\tau}$, which is the exponential decay function. This step can be seen in the section where I'm deriving $\tau$, about 3/4's the way down, in bold and caps.
If I just subtract $k_{1}$ then it won't get the correct form, but that's all I'm seeing that can be done to isolate $y$. Where did I go wrong? Thanks in advance.
Here's the same work as what's from the link:
To derive $\tau = \frac{1}{k_{1} + k_{-1}}$
Again, we start with the differential equation, but this time we'll integrate. For ease of notation, we'll first make the substitutions $O = y$ and $\frac{\mathrm{dO}}{\mathrm{d} t} = y'$.
Beginning with an intermediate step from the previous section,
$$\frac{\mathrm{dO} }{\mathrm{d} t} = k_{1} - O(k_{1} + k_{-1})$$
$$y' = k_{1} - y(k_{1} + k_{-1})$$
and now applying separation of variables,
$$\frac{y'}{k_{1} - y(k_{1} + k_{-1})} = 1$$
We then use $u$-substitution to integrate the L.H.S. of the equation, where
$u = k_{1} - y(k_{1} + k_{-1})$ and $u' = -y'(k_{1} + k_{-1})$, which yields
$$ \int \frac{y'}{k_{1} - y(k_{1} + k_{-1})} = \int 1$$
$$-\frac{1}{k_{1} + k_{-1}} \: \int \frac{u'}{u} = \int 1$$
$$-\frac{ln|u|}{k_{1} + k_{-1}} = t$$
Solving for $u$, we have
$$- \: \frac{ln|u|}{k_{1} + k_{-1}} = t$$
$$ln|u| = \: - \: t(k_{1} + k_{-1})$$
$$u = exp\begin{Bmatrix} - t \: (k_{1} + k_{-1}) \end{Bmatrix}$$
which is the same as (HERE)
$$k_{1} - y(k_{1} + k_{-1}) =exp \begin{Bmatrix} - t \: (k_{1} + k_{-1}) \end{Bmatrix}$$
Now, solving for $y$ and back-substituting $O$ for $y$, (AND HERE)
$$O = \frac{k_{1}}{k_{1} + k_{-1}} \: \: exp \begin{Bmatrix} - t \: (k_{1} + k_{-1}) \end{Bmatrix}$$
and noticing that $\frac{k_{1}}{k_{1} + k_{-1}} = O_{inf}$ we have
$$O = O_{inf} \: \: exp \begin{Bmatrix} - t \: (k_{1} + k_{-1}) \end{Bmatrix}$$
Lastly, we re-express the exponential argument as
$$- t \: (k_{1} + k_{-1}) = \frac{-t}{\frac{1}{k_{1} + k_{-1}}}$$
where $\tau = \frac{1}{k_{1} + k_{-1}},$ to then get
$$O = O_{inf} \: \: e^{-t/\tau}$$
which fits the form of the exponential decay equation, $N(t) = N_{0} \: e^{-t/\tau}$, where $N = O$ and $N_{0} = O_{inf}$, and $\tau$ is defined to be the average length of time that an ion remains unchanged by the conductance state of an ion channel at a given time $t$.
Admittedly, I have zero knowledge of ion channels. So if this makes no sense, my apologies.
That being said, after a couple minutes of Googling the subject, I believe the following holds:
1.) if $O(t)$ denotes the probability of finding an open channel when the channel opened at time $0$ (again, no idea what this means), then $O(0) = 1$;
2.) rather than $O(t) = O_{inf}e^{-t/\tau}$, it looks like the form of this function should be $$O(t)= O_{inf} + (O_0-O_{inf})e^{-t/\tau},$$ where $O_0=O(0)$, and $O_{inf} = \lim \limits_{t \to \infty} O(t)$.
(citing these two sources: Ionic Channels in Cells and Model Systems , Ion Channels and Electrical Signaling )
Given these two assumptions, I believe the misstep you've made is that you antidifferentiated without adding a constant; that is, you've integrated without regard to initial conditions.
When you have $$ -\frac{1}{k_1+k_{-1}}\int \frac{u'}{u} = \int 1$$ this is not exactly correct. Since we are integrating from time $0$ up through some time $t$, we must accommodate appropriately:
$$ -\frac{1}{k_1+k_{-1}}\int_{u(0)}^{u(t)} \frac{u'}{u} = \int_{0}^{t} 1$$
If you continue from here, you should be able to end up with the form I have mentioned above.
Hope this helps!