# Differential Equations: Finding general solution problem

1. Dec 14, 2008

### kdawghomie

1. The problem statement, all variables and given/known data

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!

2. Dec 14, 2008

### CompuChip

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.

3. Dec 14, 2008

### kdawghomie

OK, thank you. So... to clarify, I can essentially ignore the I and just work with the I's coefficient as a constant, correct?

4. Dec 15, 2008

### CompuChip

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.