Suppose $Kf=\lambda f$. Differentiating the integral expression on the two sides of $y=x$ gives
$$
(Kf)''=Kf-2f,
\qquad
(Kf)'(0)=Kf(0),
\qquad
(Kf)'(1)=-Kf(1).
$$
Therefore
$$
f''=\frac{\lambda-2}{\lambda}f=-\omega^2f,
\qquad
f'(0)=f(0),
\qquad
f'(1)=-f(1).
$$
The first boundary condition makes
$$
f(x)=A\left(\cos(\omega x)+\frac1\omega\sin(\omega x)\right).
$$
Substitution into the second gives
$$
2\cos\omega+\left(\frac1\omega-\omega\right)\sin\omega=0,
$$
or
$$
\boxed{\frac{2\omega}{\omega^2-1}=\tan\omega},
\qquad
\boxed{-\omega^2=\frac{\lambda-2}{\lambda}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 326 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-326/2/b/solution