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Consider the boundary conditions $u_x(0,y) = 0$ and $u_x(L,y) + u(L,y) = 0$.

Therefore, if $u(x,y)$ is of the form $u(x,y) = \varphi(x)\psi(y)$, $\varphi_n(x) = A\cos\lambda_nx$ and the eigenvalues are determined by

$$

\tan\lambda_n = \frac{1}{\lambda_n}.

$$

So we have that

$$

\begin{alignat*}{3}

u(x,y) & = & \sum_{n = 1}^{\infty}A\cos\lambda_nx(B\cosh\lambda_ny + C\sinh\lambda_ny)\\

& = & \sum_{n = 1}^{\infty}\cos\lambda_nx(A_n\cosh\lambda_ny + B_n\sinh\lambda_ny)

\end{alignat*}

$$

Now because of the first boundary condition, $u_y(x,0)$, we have that $B_n = 0$.

Therefore, the solution is of the form

$$

u(x,y) = \sum_{n = 1}^{\infty}A_n\cos\lambda_nx\cosh\lambda_ny.

$$

Therefore, if $u(x,y)$ is of the form $u(x,y) = \varphi(x)\psi(y)$, $\varphi_n(x) = A\cos\lambda_nx$ and the eigenvalues are determined by

$$

\tan\lambda_n = \frac{1}{\lambda_n}.

$$

So we have that

$$

\begin{alignat*}{3}

u(x,y) & = & \sum_{n = 1}^{\infty}A\cos\lambda_nx(B\cosh\lambda_ny + C\sinh\lambda_ny)\\

& = & \sum_{n = 1}^{\infty}\cos\lambda_nx(A_n\cosh\lambda_ny + B_n\sinh\lambda_ny)

\end{alignat*}

$$

Now because of the first boundary condition, $u_y(x,0)$, we have that $B_n = 0$.

Therefore, the solution is of the form

$$

u(x,y) = \sum_{n = 1}^{\infty}A_n\cos\lambda_nx\cosh\lambda_ny.

$$

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