# Further simplifiable?

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.

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$\lambda_n$ is the eigenvalue of your DE? –  Guess who it is. 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