# [SOLVED]Homogeneous System

#### dwsmith

##### Well-known member
$R(r) = A_m\mathcal{J}_m(kr) + B_m\mathcal{Y}_m(kr)$
$$\begin{pmatrix} \mathcal{J}_m(ka) & \mathcal{Y}_m(ka)\\ \mathcal{J}_m(kb) & \mathcal{Y}_m(kb) \end{pmatrix} \begin{pmatrix} A_{m}\\ B_{m} \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}$$
In order for our system to have a non-trivial solution, we require that the determinant be 0.
Therefore,
$$\mathcal{J}_m(ka)\mathcal{Y}_m(kb) - \mathcal{J}_m(kb)\mathcal{Y}_m(ka) = 0$$
which is our eigenvalue equation.

How do I get to this now:
$$R(r) = A_m\left[\mathcal{J}_m(kr)-\frac{\mathcal{J}_m(ka)}{\mathcal{Y}_m(ka)} \mathcal{Y}_m(kr)\right]$$

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