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Given the problem: $$(\kappa(x)X^{'})^{'}+\lambda\rho(x)X=0$$ for $0<x<l$ with $X(0)=X(l)=0$ where $\kappa(x)=\kappa_{1}^{2}$ for $x<a$, $\kappa(x)=\kappa_{2}^{2}$ for $\kappa>a$. $\rho(x)=\rho_{1}^{2}$ for $x<a$, and $\rho(x)=\rho_{2}^{2}$ for $x>a$. All these constants are positive and $0<a<l$

The coefficient $\kappa(x)$ and $\rho(x)$ are piecewise functions brings some troubles, how to find the equation to determine the eigenvalue?

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Your coefficients are piecewise constant, which allows you to write the general solution of your differential equation. First write the general solutions on each interval where they are constant. This will give you 4 arbitrary constants in your two solutions. Then use the matching conditions for $X$, $X'$ at the point of discontinuity ($X$ and $\kappa X'$ must be continuous on the whole interval.) This will eliminate two of the 4 arbitrary constants. This gives you an explicit general solution, depending on the two arbitrary constants. Plug the boundary conditions to it to obtain an equation for eigenvalue.

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