= Solution
For $p=0$ and $\lambda=1$, the <linear ordinary differential equation> is $y''''=y$, whose general solution is
$$
y=A\cos x+B\sin x+C\cosh x+D\sinh x.
$$
By the stated parity reduction, an <even function>[even] eigenfunction has the form $y=A\cos x+C\cosh x$. The conditions $y(c)=y'(c)=0$ have a nonzero solution exactly when
$$
\det\begin{pmatrix}
\cos c&\cosh c\\
-\sin c&\sinh c
\end{pmatrix}=0,
$$
that is,
$$
\cos c\sinh c+\sin c\cosh c=0.
$$
An <odd function>[odd] eigenfunction has the form $y=B\sin x+D\sinh x$, and the corresponding determinant gives
$$
\cos c\sinh c-\sin c\cosh c=0.
$$
Thus $\lambda=1$ is an eigenvalue precisely under one of the two stated conditions.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4-fourth-order-eigenvalue-condition.png]
{title=The fourth-order <eigenvalue> condition and its roots}
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 is
$$
\tan c=-\tanh c.
$$
There 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,
$$
\boxed{c\approx2.365020372}.
$$
\b[Hence the combined $\pm$ condition has exactly one solution in $0<c<\pi$.]
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