# 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.

• How did you get $y + k$ terms from $ky +1$ terms? I have not gone through your calculations - maybe if you could update with the integrals you performed. – Chinny84 Mar 9 '17 at 14:51
• @Chinny84: Shit. One second. – Freeman Mar 9 '17 at 14:56
• @Chinny84: That fixed it. The integral is wrong. – Freeman Mar 9 '17 at 14:59
• Been there and done it - good thing about getting old you have seen alot of short cuts for checking your result ;)! – Chinny84 Mar 9 '17 at 15:02
• @Chinny84: You sound like my supervisor xD – Freeman Mar 9 '17 at 15:11

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:

$$\ln\left(0.5945\right)+9.99885906\ln\left(0.931796325795775\right)+0.6819589257422061=\text{C}\tag9$$

• Thank you for helping! Very much appreciated, it's all sorted now! – Freeman Mar 13 '17 at 13:29