Quote:
Originally Posted by qtwerp 
x²y'' + 3xy' + 2y = 0
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No, it's a "simple" Cauchy-Euler differential equation. y' is the first derivative of y and y'', the second derivative. It should be possible without having to write down a lot of work. It's not that difficult if you've done them before.
Here's the big hint for it: y = x^m
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Thanks for the hint
m ^2 x^m + 3 m x^m + 2 x^m = 0
(x != 0)
m^2 + 3m + 2 = 0
m = -1, m = -2
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How about
:
Prove:
\sum_{n = 1}^{\infty} 1/n^s = 0 \iff \Real(s) = 1/2 where s != {-2,-4,-6...}
Nahh... just messing with you. Just state the name of the problem