The professor wrote the differential equation f²/f'=1 on the blackboard, and asked the students to solve it.

Everybody started working with the usual methods, except for a kid at the back of the class, who happened to have skipped that material, but was very bright.

Can you solve this equation without any integration?

Let's write [a0, a1, a2, a3...] as short for f(x)= a0 + a1.x + a2.x^2 + a3.x^3 + ....

Then f ' = [a1, 2.a2, 3.a3, 4.a4...] and f^2 = [a0^2, 2.a0.a1, a1^2+2.a0.a2, 2.a1.a3+2.a0.a4, ...]

Equating f ' and f^2 we get

a1 = a0^2
2.a2 = 2.a0.a1 = 2.a0^3, so a2 = a0^3
3.a3 = a1^2+2.a0.a2 = 3.a0^4, so a3 = a0^4
4.a4 = 2.a1.a3+2.a0.a4 = 4.a0^5, so a4 = a0^5

and then f=[a0, a0^2, a0^3, a0^4...]

Writing just "a" instead of a0, f(x)=a/(1-ax).

Posted by e.g. on 2006-05-04 09:33:36