Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an equation $(xy')'+kxy=xf(x)$ where $k$ is not an eigenvalue; $y(x)$ and $f(x)$ are subjected to boundary conditions (i) bounded as $x\to 0$; (ii) $y(1)=0=f(1)$

I want to get the (unique) solution $y(x)$ as a linear combination of the eigenfunctions $J_0(z_n x)$ where $z_n$ are the zeros of $J_0$.

So I want $y(x)=\sum a_nJ_0(z_nx)$. Am I right to think that $a_n={\int_0^1J_0(z_nx)f(x)dx\over \lambda_n+k}$? What more can I conclude? Can I further simplify the expression?

Added: $\lambda_n$ are eigenvalues of the system.

Thanks.

share|improve this question
    
$\lambda_n$ is the eigenvalue of your DE? –  J. M. Oct 21 '11 at 13:06
    
@J.M.: Indeed.. –  kristoff Oct 21 '11 at 13:08
    
@kristoff: Would you care to include this information in your question? –  Rasmus Oct 21 '11 at 13:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.