Hi guys I am doing some differential equations and I got this one:enter image description here

I have no idea how he went from (38) to (39). When I solve it by integrating factor, the equation I have is: $$c_{A}{(t)}=\frac{{c_{af}}}{k\tau+1}+c_{A0}e^{-(\frac{1}{\tau}+k)t}$$ however they have additional $e^{blabla}$

It is this website:http://jbrwww.che.wisc.edu/home/jbraw/chemreacfun/ch4/slides-matbal.pdf Page 57 as you can see. Am I missing something?


We have

$$c_A'(t)=\frac1\tau (c_{Af}-c_A(t))-kc_A(t)=\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(t)$$

We can therefore write

$$t=\int_0^t\frac{1}{\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(t')}\,dc_A(t') \tag 1$$

Evaluating the integral in $(1)$ reveals

$$t=\frac{\log\left(\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(0)\right)-\log\left(\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(t)\right)}{k+\frac1\tau} \tag 2$$

Solving $(2)$ for $c_A(t)$ yields

$$\begin{align} c_A(t)&=\frac{\frac1\tau c_{Af}-\left(\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(0)\right)e^{-\left(k+\frac1\tau\right)t}}{k+\frac1\tau}\\\\ &=c_{A}(0)e^{-\left(k+\frac1\tau\right)t}+\frac{c_{Af}}{k\tau +1}\left(1-e^{-\left(k+\frac1\tau\right)t}\right)\\\\ &=c_{A0}e^{-\left(k+\frac1\tau\right)t}+\frac{c_{Af}}{k\tau +1}\left(1-e^{-\left(k+\frac1\tau\right)t}\right) \end{align}$$

as was to be shown!

  • $\begingroup$ Thank you for this. Can you explain to me why the simple integrating factor method that I used failed? $\endgroup$ – Scavenger23 Jan 30 '16 at 16:42
  • $\begingroup$ You're welcome. My pleasure. - Mark $\endgroup$ – Mark Viola Jan 30 '16 at 16:46
  • $\begingroup$ What was your integrating factor? $\endgroup$ – Mark Viola Jan 30 '16 at 16:46
  • $\begingroup$ $\frac{1}{\tau}+k$ integrating that we get $exp(\frac{1}{\tau}+k)t $ $\endgroup$ – Scavenger23 Jan 30 '16 at 16:50

Well, your simple integrating factor method actually works. You just have to determine the constant correctly with the initial condition: $c_A(0)=c_{A0}=\frac{c_{af}}{k\tau+1}+Ke^0$, now solve for $K$

  • $\begingroup$ i see :) i chose constant wrong. Thanks. $\endgroup$ – Scavenger23 Jan 30 '16 at 16:51

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