1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $K$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.