What Are the Eigenvalues and Eigenfunctions of the Operators x d/dx and d/dx x?

[eit32]

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]

Write down the eigenvalue equation. This will be a differential equation, which you will need to solve..

 

[eit32]

i don't think i quite understand what you are saying

 

[CompuChip]

What does it mean for a function f to be an eigenvalue of x(d/dx)? What equation must it satisfy?

 

[eit32]

if L is an opertaor than
L[f(x)]=cf(x); where f(x) is an eigenfunction and c is an eigenvalue

 

[CompuChip]

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]

x d/dx[f(x)]=cf(x)...not really

 

[CompuChip]

OK, hint: try f(x) = x^k for some k

 

[kdv]

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]

Why "not really"? This is correct!



x y' = c x

I think you meant

<br /> x y&#039; = c y<br />

 

[kdv]

I think you meant

<br /> x y&#039; = c y<br />

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]

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]

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]

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.