Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 17B / c / Solution

For $p=0$ and $\lambda =1$, the linear ordinary differential equation is $y^{\prime \prime \prime \prime }=y$, whose general solution isBy the stated parity reduction, an even eigenfunction has the form $y=A\cos x+C\cosh x$. The conditions $y(c)=y^{\prime }(c)=0$ have a nonzero solution exactly whenthat is,An odd eigenfunction has the form $y=B\sin x+D\sinh x$, and the corresponding determinant givesThus $\lambda =1$ is an eigenvalue precisely under one of the two stated conditions.

For the minus sign, division by $\cos c\cosh c$ away from $c=\pi /2$ gives $\tan c=\tanh c$. On $(0,\pi /2)$, $\tan c>\tanh c$ for $c>0$, while on $(\pi /2,\pi )$ their signs differ, so there is no root. For the plus sign the equation isThere is no root in $(0,\pi /2)$. On $(\pi /2,\pi )$, the function $\tan c+\tanh c$ increases strictly from $-\infty$ to $\tanh \pi >0$, so it has exactly one root by the intermediate value theorem. Numerically,Hence the combined $\pm$ condition has exactly one solution in $0<c<\pi$.