MHB Please check Laguerre's eqtn solution

[ognik]

Hi - revising Sturm-Liouville theory and would appreciate someone checking & correcting/improving the following.

Given Laguerre's eqtn $xy''+(1-x)y'+\lambda y = 0 $, which is not self-adjoint, find the weighting factor w(x).

If I didn't know the integrating factor, would one be $ \mu(x) = \frac{1}{x} e^{\int\frac{1-x}{x}dx} = \frac{1}{x}\left( x-e^{x} \right)$? As it is I know $\mu(x) = e^{-x}$ works well.

Then $ e^{-x} xy''+ e^{-x} (1-x)y'+ e^{-x} \lambda y = 0 $ and I verify this is self-adjoint by showing $p'_0$ = $p_1$

Then by comparison with the Sturm-Liouville problem $ \mathcal{L}y + \lambda W(x)y = 0 $, I can write:
$q(x) = 0$
$\lambda$ is the eigenvalue and y(x) is the corresponding eigenfunction
and $w(x) = e^{-x}$

Anything I am missing?

 

[ognik]

Hi - the reason I ask questions like this is that over this year some things I thought were right have been - to some degree - wrong or incomplete or just using incorrect notation. So if I have this correct I would really appreciate a short confirmation; but am also interested in my 1st question about the Integration Factor - it would seem there can be more than 1 I/F ?
Thanks.

 

[Mathwebster]

. . . $ \mu(x) = \frac{1}{x} e^{\int\frac{1-x}{x}dx} = \frac{1}{x}\left( x-e^{x} \right)$?

You might want to check what you did after you integrated.

 

[ognik]

Sorry, this was ambiguous. After I integrated I didn't use the IF that I had found, because I had used an IF of $e^{-x}$ in a different Laguerre's eqtn problem and it looked easier to work with than the IF I found by integrating.

I just want to confirm that there can be multiple different IFs for a function - i.e. integrating is just one method to find 1 possibility?

Secondly, using the easier IF of $e^{-x}$, I just wanted to check that I had applied the S-L theory correctly (book doesn't give the answer)

Thanks

 

[Mathwebster]

What I'm saying is that after your integration, your simplifaction is incorrect.

$$\dfrac{e^{\int\frac{1-x}{x}dx}}{x} = \dfrac{e^{ln x - x}}{x} = \dfrac{x e^{-x}}{x} = e^{-x}$$

(I've also be a little relaxed with absolutes).

 

[ognik]

Oops misunderstood, you're of course right. So is there then only 1 IF that will make a function self-adjoint?

And please check that the rest of what I did is correct?