Integrating factor vs. Laplace. Engineering problems

Preferred solving method

  • Laplace transform

    Votes: 0 0.0%

  • Total voters
    3
  • #1
232
53
Hello PF! We were doing mass balances on mixing tanks in one of my ChemE courses, and in one of the problems we arrived at the following DE:
[tex]\frac{dC_B}{d \theta} + 0.025C_B=0.0125 e^{-0.025 \theta}[/tex]
Where CB is the concetration of salt in the tank and θ is time. The professor made us solve the equation using two methods, integrating factor and Laplace transform, and told us to keep working on following problems with the method of our choice. Personally, I preferred using the integrating factor method, because it is a very simple equation, and I don't like to be dealing with transform charts while solving problems, although I know that for more complex DEs the integrating factor method becomes really complicated, and Laplace transform remains relatively simpler. I think I'll be using integrating factor for simple equations like this, and LT for more complex differential equations, although I'm more inclined to use the integrating factor method. Also, which method do you think looks more elegant? Even though I'm an engineering student I always try to be "mathematically elegant," whenever it is possible. By the way, the solution of the equation is:
[tex]C_B= (0.0125 \theta +0.25) e^{-0.025 \theta}[/tex]
So, long story short, integrating factor or Laplace transform? Why? Which method do you think is more elegant?

Thanks in advance for any input!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
In my opinion, while both would give you the same solution. for your specific equation, I'd use the integrating factor method.

Mainly because I'd think the Laplace transform is kind of overkill for simple DEs.
 
  • Like
Likes LAZYANGEL and MexChemE
  • #3
15
1
Integrating factor sounds way easier, because the integrating factor eliminates the exponential on the right hand side which is fantastic.

##\mu (x)=e^{\int 0.025 \: d\theta}=e^{0.025 \theta}##

And when you multiply both sides by the integrating factor, you get:

## \frac{d}{d \theta} \left [ C_B \cdot e^{0.025\theta} \right ] = 0.0125##

##\int \frac{d}{d \theta} \left [ C_B \cdot e^{0.025\theta} \right ] \: d \theta =\int 0.0125 \: d \theta##

##C_B \cdot e^{0.025\theta}=0.0125 \theta + c_1##

##C_B=e^{-0.025\theta}\left ( 0.0125 \theta + c_1 \right )##
 
  • Like
Likes MexChemE

Related Threads on Integrating factor vs. Laplace. Engineering problems

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
3K
Replies
7
Views
21K
Replies
7
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
6K
Top