How to solve $\frac{\text{d}y}{\text{d}\tau} = -\varepsilon y/\left(1+\frac{k\Delta}{(ky+1)^2}\right)$ If we take
$$\frac{\text{d}y}{\text{d}\tau} = -\varepsilon y/\left(1+\frac{k\Delta}{(ky+1)^2}\right) \qquad\qquad (1)$$
with $y(0)=y_0$ then we can solve it as such:
\begin{align}
-\varepsilon\tau + C_0
&=
\int
\frac
{1+\frac{k\Delta}{\left(ky+1\right)^2}}
{y}
\ \text{d}y
\\
%%
&=
\int
\frac
{1}
{y}
%
+
%
\frac
{k\Delta}
{y\left(ky+1\right)^2}
\ \text{d}y\\
%%
&=
\log{y}
%
+
%
\frac{\Delta}{k}\log{\left(\frac{y}{k+y}\right)}
%
+
%
\frac{\Delta}{k+y}
\end{align}
with
\begin{align}
C_0
&=
\log{y_0}
%
+
%
\frac{\Delta}{k}\log{\left(\frac{y_0}{k+y_0}\right)}
%
+
%
\frac{\Delta}{k+y_0}
\end{align}
Therefore we can write:
\begin{align}
\tau=
-
\frac{1}{\varepsilon}
\left(
\log{y}
%
+
%
\frac{\Delta}{k}\log{\left(\frac{y}{k+y}\right)}
%
+
%
\frac{\Delta}{k+y}
-C_0\right) \qquad\qquad (2)
\end{align}
I have solved $(1)$ numerically, using an ODE solver in MATLAB, with the following parameters:
$$\Delta=0.4351$$ 
$$\varepsilon=4.4261\times 10^{-5}$$
$$k=22.9806$$ 
$$y_0=0.5945$$
$$\tau\in [0,200000]$$
This is plotted by the black line in the following figure. I then solved $(2)$ for $\tau$ for $y\in[y_\text{min},y_0]$ where $y_\text{min}$ was the smallest value for y output from the numerical simulation. I then plotted this on the same figure, in dashed magenta. They don't match! I have checked this over and over, I do not know what I have done wrong here. Any help would be really appreciated.

 A: Well, as you said we can solve it:
$$\text{y}'\left(\tau\right)=-\frac{\epsilon\cdot\text{y}\left(\tau\right)}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{y}\left(\tau\right)\right)^2}}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(\tau\right)}{\frac{\text{y}\left(\tau\right)}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{y}\left(\tau\right)\right)^2}}}\space\text{d}\tau=\int-\epsilon\space\text{d}\tau\tag1$$
Now, we get:


*

*$$\int-\epsilon\space\text{d}\tau=-\epsilon\int1\space\text{d}\tau=\text{C}_1-\epsilon\cdot\tau\tag2$$

*Substiute $\text{u}=\text{y}\left(\tau\right)$:
$$\int\frac{\text{y}'\left(\tau\right)}{\frac{\text{y}\left(\tau\right)}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{y}\left(\tau\right)\right)^2}}}\space\text{d}\tau=\int\frac{1}{\frac{\text{u}}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{u}\right)^2}}}\space\text{d}\text{u}=\int\frac{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{u}\right)^2}}{\text{u}}\space\text{d}\text{u}=$$
$$\int\frac{1}{\text{u}}\space\text{d}\text{u}+\text{k}\cdot\Delta\int\frac{1}{\left(1+\text{k}\cdot\text{u}\right)^2}\cdot\frac{1}{\text{u}}\space\text{d}\text{u}\tag3$$

*$$\int\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|\text{u}\right|+\text{C}_2\tag4$$

*$$\int\frac{1}{\left(1+\text{k}\cdot\text{u}\right)^2}\cdot\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|-\text{k}\cdot\text{u}\right|-\ln\left|1+\text{k}\cdot\text{u}\right|+\frac{1}{1+\text{k}\cdot\text{u}}+\text{C}_3\tag5$$


So, we get:
$$\ln\left|\text{y}\left(\tau\right)\right|+\text{k}\cdot\Delta\cdot\left\{\ln\left|-\text{k}\cdot\text{y}\left(\tau\right)\right|-\ln\left|1+\text{k}\cdot\text{y}\left(\tau\right)\right|+\frac{1}{1+\text{k}\cdot\text{y}\left(\tau\right)}\right\}=\text{C}-\epsilon\cdot\tau\tag6$$
We can simplify a very little:


*

*$$\left|-\text{k}\cdot\text{y}\left(\tau\right)\right|=\left|-\text{k}\right|\cdot\left|\text{y}\left(\tau\right)\right|=\left|\text{k}\right|\cdot\left|\text{y}\left(\tau\right)\right|\tag7$$

*$$\ln\left|-\text{k}\cdot\text{y}\left(\tau\right)\right|-\ln\left|1+\text{k}\cdot\text{y}\left(\tau\right)\right|=\ln\left(\frac{\left|\text{k}\right|\cdot\left|\text{y}\left(\tau\right)\right|}{\left|1+\text{k}\cdot\text{y}\left(\tau\right)\right|}\right)\tag8$$



So, when we use the values:
$$\ln\left(0.5945\right)+9.99885906\ln\left(0.931796325795775\right)+0.6819589257422061=\text{C}\tag9$$
