Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have been simply asked to integrate the following: $dc/dt = 1-c-a c$.

I have used the integrating factor as $e^{(1+a) t}$.

Then finish with $c=1/(1+a)$, however my lecturer gets $c=\frac{1}{1+a}(1-e^{(a+1)t})$.

Could anyone at all show me the method to get this. I know it is meant to be a relatively easy question, but i just don't know what i am doing wrong.

Thank you.

share|improve this question
    
Separate variables. –  David Mitra May 9 '12 at 15:47
1  
Is $e$ a constant or a function of $(1 + a) t$? –  tentaclenorm May 9 '12 at 15:48
    
Is $*$ a convolution? (Only computer languages use $*$ for multiply.) –  GEdgar May 9 '12 at 15:54
    
sorry when i typed it the computer changed things, * is just multiplication. When i use the method of an integrating factor, my integrating factor is exp((1+a)t), so not e the constant. I thought i couldn't use separation of variables for this type of problem? –  R.M May 9 '12 at 15:59

3 Answers 3

up vote 2 down vote accepted

I don't think integrating factors are needed here.

Note that $$ \begin{align} t &=\int\frac{\mathrm{d}c}{1-(a+1)c}\\ &=-\frac{1}{a+1}\log(1-(a+1)c)+t_0\tag{1} \end{align} $$ Solving $(1)$ yields $$ \begin{align} c &=\frac{1}{a+1}\left(1-e^{-(a+1)(t-t_0)}\right)\\ &=\frac{1}{a+1}-\left(\frac{1}{a+1}-c_0\right)e^{-(a+1)t}\tag{2} \end{align} $$ $c=\dfrac{1}{a+1}$ is a particular solution of $(2)$ with $c_0=\frac{1}{a+1}$ ($t_0=-\infty$),

share|improve this answer
    
ahhhhh thank you, in yours you have $t-t_o$ whereas he just has t, are you allowed to just put t there? –  R.M May 9 '12 at 16:03
    
@R.M: $t_0$ is the constant of integration. It can be anything, $0$ is fine. Since we could have written $e^{-(a+1)(t-t_0)}=Ce^{-(a+1)t}$, we could set $C=0$ which would be the same as setting $t_0=-\infty$; this is why $t_0=-\infty$ is valid. –  robjohn May 9 '12 at 16:10
    
@R.M you can determine the explicit value of $t_0$ if you are given an initial condition for your problem. Presumably, your lecturer used $c(0) = 0$ to conclude that $t_0 = 0$. –  tentaclenorm May 9 '12 at 16:12
    
Thank you so much everyone!!!! –  R.M May 9 '12 at 16:15

$$ \frac{dc}{dt}+(1+a)c = 1 $$ multiply by integrating factor $e^{(1+a)t}$: $$\begin{align} &e^{(1+a)t}\frac{dc}{dt}+(1+a)e^{(1+a)t}c=e^{(1+a)t} \\ &\frac{d}{dt}\left(e^{(1+a)t}c\right)=e^{(1+a)t} \\ &e^{(1+a)t}c = \int e^{(1+a)t}\,dt = \frac{e^{(1+a)t}}{1+a}+A \\ &c = \frac{1}{1+a}+Ae^{-(1+a)t} \end{align}$$ for some constant $A$. Presumably there is an initial condition (that R.M. has not told us) to let the lecturer determine the constant $A$

share|improve this answer
    
no we aren't given any IC's either which is why i'm so confused by it all. but thanks so much!!! –  R.M May 9 '12 at 16:11
    
Although I was able to solve this without integrating factors, if the right side of your first equation were not a function of $c$, an integrating factor would most likely be necessary. (+1) –  robjohn May 9 '12 at 16:54

what @robjohn said is true:

because of the nature of this equation it is separable so an integrating factor is not strictly necessary. but because you requested it be done in that method:

first consider the form of the equation for an integrating factor:
A differential equation of the form $y' + P(x)y = Q(x) $ can be solved by
$ y = \frac{\int{\mu Q(x)} dx}{\mu} + \frac{k}{\mu} $ where $k$ is an arbitrary constant and $\mu $ is the integrating factor ($\mu = e^{\int{P(x)}dx}$)

Let's fit your equation into the form needed for the method of integrating factors (In your case, $y$ is $c$ ):

$ c' + c(1+a) = 1 $

so you should obtain for an integrating factor of $ \mu = e^{\int{(1+a)}dt} = e^{t(1+a)} $

so $ c = \frac{1}{e^{t(1+a)}} \int{(e^{t(1+a)})dt} + \frac{k}{e^{t(1+a)}}$

if you integrated correctly, you should get $\int{(e^{t(1+a)})dt} = \frac{1}{1+a}e^{t(1+a)}$ which you will find nicely cancels with the integrating factor, $\mu$ which is already in the denominator. So your final answer should be

$ c = \frac{1}{1+a} + \frac{k}{e^{t(1+a)}} = \frac{1}{1+a} + ke^{-t(1+a)} $

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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