这是一份kcl伦敦大学学院 6CCM318A作业代写的成功案
The general solution of $y^{\prime \prime}+\lambda y=0$ is
$y=A \cos \sqrt{\lambda} x+B \sin \sqrt{\lambda} x$
Then from the condition $y(0)=0$ we find $A=0$, so that
$$
y=B \sin \sqrt{\lambda} x
$$
The condition $y^{\prime}(L)+\beta y(L)=0$ gives
$$
B \sqrt{\lambda} \cos \sqrt{\lambda} L+\beta B \sin \sqrt{\lambda} L=0{ }^{\prime} \quad \text { or } \quad \tan \sqrt{\lambda} L=-\frac{\sqrt{\lambda}}{\beta}
$$
6CCM318ACOURSE NOTES ：
$$
R=r^{-1 / 2}\left[A r \sqrt{1 / 4-\lambda^{2}}+B r^{\left.-\sqrt{1 / 4-\lambda^{2}}\right]}\right.
$$
or
$$
R=A r^{-1 / 2}+\sqrt{1 / 4-\lambda^{2}}+B r^{-1 / 2-\sqrt{1 / 4-\lambda^{2}}}
$$
This solution can be simplified if we write
$$
-\frac{1}{2}+\sqrt{\frac{1}{4}-\lambda^{2}}=n
$$
so that
$$
-\frac{1}{2}-\sqrt{\frac{1}{4}-\lambda^{2}}=-n-1
$$
In sueh case $(1)$ becomes
$$
R=A r^{n}+\frac{B}{r^{m+1}}
$$
Multiplying equations $(2)$ and $(s)$ together leads to
$$
\lambda^{2}=-n(n+1)
$$