Differential Equations: Finding general solution problem

kdawghomie
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Homework Statement



Find the general solution of (D^4 - I)^2(D^2 - 4D + 13I)^2(y) = 0

2. The attempt at a solution

My issue with this problem is that I have no clue as to what the I's mean. I have become familiar with D being used notationally with differential equations, but the introduction of the I's is totally foreign to me, and my professor has never even addressed them. Am I supposed to assume the I's are simply a constant, or am I totally missing something here?

I can break the problem down to the following, though:
[(D^2 + sqrt(I))(D + sqrt(I))(D - sqrt(I))]^2 * [D^2 - 4D +13I]^2 * y = 0

From here, am I supposed to proceed "as usual" with solving the equation... or do the I's have some significance? It seems very possible to solve w/ I's being a constant, but absolutely brutal to actually find the gen. solution for :/

Thank you for any help!
 
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I is usually used for the identity operator: I f = f.
So for example, solving
(D^2 - 3 I) = 0
would mean: find all functions such that
\frac{d^2}{dx^2} f(x) - 3 f(x) = 0.

Also used is "id", a 1 with a hat, or a double struck 1.
Physicists tend to leave out the I altogether, and write
(D^2 - 3) = 0
for the above equation.
 
OK, thank you. So... to clarify, I can essentially ignore the I and just work with the I's coefficient as a constant, correct?
 
Essentially, yes.
Note however, that I is an operator, just like D is, so you cannot really write sqrt(I) any more than you can sqrt(D). You can use that I^n = I though, for any (integer) n.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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