How would you solve the differential equation

with the initial condition ? I know what the answer is supposed to be, but I don’t know how to directly solve it.

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

(to make a partition of , you can put anywhere from to elements in the same set with ; there are ways to choose the elements to include, and ways to partition the rest).

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About Brent

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

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?

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.

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

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

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.

Cool, thanks.

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).

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}

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

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

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

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

∫ f’/f dx = log f

Separation of variables

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

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