How to solve this differential equation?

Posted on December 16, 2009 by Brent

How would you solve the differential equation

$B'(x) = e^x B(x)$

with the initial condition $B(0) = 1$ ? I know what the answer is supposed to be, but I don’t know how to directly solve it.

In case you’re wondering, $B(x)$ is the exponential generating function for the Bell numbers, which count set partitions. The differential equation in question arises from noting that

$\displaystyle B_{n+1} = \sum_{k=0}^n \binom n k B_{n-k}$

(to make a partition of $\{1, \dots, n+1\}$ , you can put anywhere from $k = 0$ to $n$ elements in the same set with $n+1$ ; there are $\binom n k$ ways to choose the $k$ elements to include, and $B_{n-k}$ ways to partition the rest).

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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This entry was posted in combinatorics, math and tagged Bell numbers, differential, equation, generating function, partitions. Bookmark the permalink.

15 Responses to How to solve this differential equation?

Mark Dominus says:

December 16, 2009 at 10:25 am

Doesn’t B(x) = c * exp(exp(x)) solve the equation, and then satisfying the initial condition is simple, you take c = 1/e? Or did I make some ridiculously simple error?

Reply
- Brent says:
  
  December 16, 2009 at 10:38 am
  
  Yes, that is the correct solution, but how did you arrive at it? Did you just stare at the equation and c * exp(exp(x)) popped into your head? I’m looking for a general method whereby one could solve this equation without having a magical flash of inspiration.
  
  Reply
  - Mark Dominus says:
    
    December 16, 2009 at 10:47 am
    
    I wouldn’t call it a magical flash of inspiration. But d/dx exp(f) is f’ exp(f).
    
    Reply
    - Brent says:
      
      December 16, 2009 at 10:51 am
      
      Fair enough. One man’s skill is another man’s magical flash of inspiration, I suppose. =)
      
      Reply
      - Mark Dominus says:
        
        December 16, 2009 at 10:58 am
        
        It’s probably just that I’ve already solved six hundred similar-looking problems by now.
        
        By the way, George Simmons’ textbook about differential equations is a great read. I was very disappointed by differential equations when I took them as a class. The Simmons book fixed that.
        
        I’ll send you the review I wrote of it a few years back.
      - Brent says:
        
        December 16, 2009 at 11:04 am
        
        Cool, thanks.
zygoloid says:

December 16, 2009 at 10:39 am

Substitute y = e^x.
Then dy/dx = e^x = y.

dB/dx = e^x B = y B
=> dB/dy = dB/dx * dx/dy = y B * dx/dy = B
=> B = k e^y = k e^(e^x)

B|x=0 = 1 => k = 1 => B = e^(e^x).

Reply
Xaositect says:

December 16, 2009 at 10:43 am

B’ = e^x B
\frac{B’}{B} = e^x
(\ln B)’ = e^x
\ln B = e^x + c
B = C e^{e^x}
B(0) = Ce => C = \frac{1}{e}

Reply
zygoloid says:

December 16, 2009 at 10:45 am

D’oh. Of course, I meant k = 1/e. B = e^(e^x – 1).

Reply
Brent says:

December 16, 2009 at 10:45 am

Ah! Thanks, I knew it was probably easy. I’ve forgotten way too much continuous math…

Reply
Gregory Wright says:

December 16, 2009 at 10:58 am

You already know how to solve this equation, but it is perhaps the presentation is slightly unusual. Divide both sides by B to get

d log B / dx = e^x

You can do this integral immediately. Looking at the
solution you’ll be able to verify that division by B
was justified (no singularities). The solution is

log B = e^x + C

Reply
Pseudonym says:

December 16, 2009 at 9:20 pm

The way I would have solved it is using Gregory Wright’s approach. I usually remember it this way:

∫ f’/f dx = log f

Reply
Heinrich Apfelmus says:

December 17, 2009 at 7:16 am

Separation of variables

Reply
jonas says:

December 26, 2009 at 11:10 am

Can’t you use that general formula for single-variable linear ODEs? The one with the exp of an integral.

Reply
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