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eit32
Mar27-08, 02:01 PM
a) Consider the operator x d/dx(where 1st d/dx acts on the function, then x acts on the resulting function by simply multiplying by x )acting on the set of functions of a real variable x for x>0. What are the eigenvalues and the corresponding eigenfunctions of this operator?

b) What about d/dx x (where 1st x acts on the function, then d/dx acts upon the resulting function)? What are the eigenvaules and corresponding eigenfunctions for this?

CompuChip
Mar27-08, 02:38 PM
Write down the eigenvalue equation. This will be a differential equation, which you will need to solve..

eit32
Mar27-08, 03:14 PM
i don't think i quite understand what you are saying

CompuChip
Mar27-08, 03:16 PM
What does it mean for a function f to be an eigenvalue of x(d/dx)? What equation must it satisfy?

eit32
Mar27-08, 03:39 PM
if L is an opertaor than
L[f(x)]=cf(x); where f(x) is an eigenfunction and c is an eigenvalue

CompuChip
Mar27-08, 03:56 PM
Exactly. What equation do you get when you plug in the L you are given.
Do you have any idea how to solve this?

eit32
Mar27-08, 04:10 PM
x d/dx[f(x)]=cf(x)....not really

CompuChip
Mar27-08, 04:44 PM
OK, hint: try f(x) = x^k for some k

kdv
Mar27-08, 04:51 PM
x d/dx[f(x)]=cf(x)....not really

Why "not really"? This is correct!

In other words,

x y' = c y

where a prime indicates a derivative with respect to x and c is some constant. This is a simple differential equation, right? Can you solve it?

kamerling
Mar27-08, 05:44 PM
Why "not really"? This is correct!



x y' = c x


I think you meant


x y' = c y

kdv
Mar27-08, 06:51 PM
I think you meant


x y' = c y


Yes indeed. Thanks for pointing out the typo. I will correct this in my post so that it does not confuse the OP.

Thanks!

eit32
Mar27-08, 10:15 PM
yeah, i took D.E. like 3 years ago and i'm a bit hazy on the details...it looks familar but i'm not really sure where to start it

kdv
Mar27-08, 10:24 PM
yeah, i took D.E. like 3 years ago and i'm a bit hazy on the details...it looks familar but i'm not really sure where to start it

It's a separable DE....

x \frac{dy}{dx} = c y \rightarrow \frac{dy}{y} = c \frac{dx}{x}

You only need to integrate both sides

merryjman
Mar29-08, 11:37 AM
Considering the quantity and type of posts you've put up at this site, eit32, you may want to dust off your DEq and linear algebra books, because most of your questions deal more with the mechanics of math operations rather than physical principles themselves.