How would I solve this :

$-(xu')'=\frac{1}{x} \ln{x} \enspace 1<x<e$

$u(1)=0 \enspace, u'(e)=0$

I want to expand this with:

$u(x) = \sum_{n=1} ^{\infty} b_n X_n(x)$

Where $X_n(x)$ are the eigenfunctions of a S-L problem.

$b_n$ are some constants.

I want to express the solution as a single integral, from x = 1 to x = e.

Can someone please show me how to expand this? How can I find the eigenfunctions and the b_n. I know it is easy to solve with different methods - but I would like to do it this way to get a better grasp of similar problems. Thanks!

The lack of a lambda is really throwing me off - no idea how to do this.


can't you integrate this straight? $$xu' = \int_x^e \frac{\ln t}{t} \, dt =(\ln t)^2\big|_x^e=1 - (\ln x)^2 \to u' = \frac 1 x- \frac{(\ln x)^2}{x} \to u = \ln x - \frac 13 (\ln)^3.$$

  • $\begingroup$ Ya sure, it is an easy DE to solve, but the point is to do with eigenfunctinos. $\endgroup$ – sci-guy Mar 16 '15 at 18:25

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