## How to solve this differential equation?

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).

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### 15 Responses to How to solve this differential equation?

1. Mark Dominus says:

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?

• Brent says:

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.

• Mark Dominus says:

I wouldn’t call it a magical flash of inspiration. But d/dx exp(f) is f’ exp(f).

• Brent says:

Fair enough. One man’s skill is another man’s magical flash of inspiration, I suppose. =)

• Mark Dominus says:

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:

Cool, thanks.

2. zygoloid says:

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).

3. Xaositect says:

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}

4. zygoloid says:

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

5. Brent says:

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

6. Gregory Wright says:

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

7. Pseudonym says:

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

∫ f’/f dx = log f

8. Heinrich Apfelmus says:
9. jonas says:

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