Integrating Factor Homework: Solving Diff. Eq w/ Constants

[gfd43tg]

Homework Statement

$C_{B}$ is a function of $\tau'$, and $k_{1}$,$k_{2}$, and $C_{A0}$ are constants. I want to solve this differential equation
\frac {dC_{B}}{d \tau'} + k_{2}C_{B} = k_{1}C_{A0}e^{-k_{1} \tau'}

Homework Equations

The Attempt at a Solution

Using the integrating factor, we get
\frac {d(C_{B}e^{k_{2} \tau&#039;})}{d \tau&#039;} = k_{1}C_{A0}e^{(k_{2} - k_{1})<br /> \tau&#039;}

However, from here I am unsure how to separate this. I was thinking of using chain rule on the inside, but that seems to only get me back to where I started (that's the purpose of the integrating factor??)

 

[pasmith]

The next step is to integrate both sides with respect to \tau&#039;. By the fundamental theorem of calculus, the integral of the left hand side is
\int \frac{d}{d\tau&#039;} (C_Be^{k_2 \tau&#039;})\,d\tau&#039; = C_Be^{k_2 \tau&#039;} + \mbox{constant}.

 

[HallsofIvy]

I don't know what you mean by "separate" or what it is you want to "separate". There is no need to "separate" anything- just go ahead and integrate:
\frac{d C_Be^{k_2\tau&#039;}}{d\tau&#039;}= k_1C_{A0}e^{(k_2- k_1)\tau&#039;}
\int d C_Be^{k_2\tau&#039;}= \int k_1C_{A0}e^{(k_2- k_1)\tau&#039;} d\tau&#039;

C_Be^{k_2\tau&#039;}= \frac{k_1C_{A0}}{k_2- k_1}e^{(k_2- k_1)}\tau&#039;+ C

That's the whole point of an "integrating factor".

 

[gfd43tg]

C_{B} e^{k_{2} \tau&#039;} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau&#039;} + C
My boundary condition is $C_{B} = 0$ at $\tau' = 0$, this means $C = -\frac {k_{1}C_{A0}}{k_{2} - k_{1}}$

Then the solution is
C_{B} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau&#039;} - \frac {k_{1}C_{A0}}{k_{2} - k_{1}}

But the solution given is
C_{B} = k_{1}C_{A0} \Big( \frac {e^{-k_{1} \tau&#039;} - e^{-k_{2}<br /> \tau&#039;}}{k_{2} - k_{1}} \Big)

 

[pasmith]

C_{B} e^{k_{2} \tau&#039;} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau&#039;} + C
My boundary condition is $C_{B} = 0$ at $\tau' = 0$, this means $C = -\frac {k_{1}C_{A0}}{k_{2} - k_{1}}$

Then the solution is
C_{B} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau&#039;} - \frac {k_{1}C_{A0}}{k_{2} - k_{1}}

No, so far you have <br /> C_B e^{k_2 \tau&#039;} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau&#039;} - \frac {k_{1}C_{A0}}{k_{2} - k_{1}}.<br />
Dividing both sides by e^{k_2 \tau&#039;} yields the given solution.

 

[gfd43tg]

Woops, I see where I made my mistake. Thank you!