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\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L + (\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0

$$

where $\phi_{n,m}$ and $\lambda_{n,m}$ represent distinct modal eigenfunctions which satisfy mixed boundary conditions at $x = 0,L$ of the form

\begin{alignat*}{3}

a\phi(0) + b\phi'(0) & = & 0\\

c\phi(L) + d\phi'(L) & = & 0

\end{alignat*}

Show that the eigenfunctions are orthogonal.

$$

\int_0^L\phi_m\phi_m dx = 0.

$$

I not sure how to proceed since there are constants a,b,c,d.

If they weren't there, I would proceed as

$$

(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = -\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L

$$

\begin{alignat*}{3}

\phi(0) &= &-\phi'(0) \\

\phi(L) &=& -\phi'(L)

\end{alignat*}

Therefore,

$$

(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0\iff \int_0^L\phi_n\phi_m dx = 0

$$