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