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\varphi(x) = A\cos x\sqrt{\lambda} + B\frac{\sin x\sqrt{\lambda}}{\sqrt{\lambda}}

$$

$$

\text{B.C.} = \begin{cases} u(0,t) = u(L,t)\\u_x(0,t) = u_x(L,t)\end{cases}

$$

Which leads to

$$

A(1 - \cos L\sqrt{\lambda}) = B\frac{\sin L\sqrt{\lambda}}{\sqrt{\lambda}} \quad \text{and} \quad B(1 - \cos L\sqrt{\lambda}) = -A\sqrt{\lambda}\sin L\sqrt{\lambda}

$$

I solved for B and obtained

$$

B = \frac{-A\sqrt{\lambda}\sin L\sqrt{\lambda}}{1 - \cos L\sqrt{\lambda}}.

$$

From this, I can substitute and get

$$

A\left[(1 - \cos L\sqrt{\lambda})^2 - \sin^2 L\sqrt{\lambda}\right] = 0

$$

So can I conclude A = 0? I am having a struggle to find A and B here so I can get the eigenfunction.