Solution. For \(\lambda \leq 0\) only the zero solution satisfies both conditions. For \(\lambda = k^2 > 0\), \(y = \sin kt\) from the first condition, and \(y'(1) = k\cos k = 0\) gives \(k = (n - \frac 12)\pi \). So \(\lambda _n = (n - \frac 12)^2\pi ^2\) and \(\phi _n = \sin (n - \frac 12)\pi t\), \(n = 1, 2, \dots \). For \(m \neq n\), \[ \int _0^1\sin (m - \tfrac 12)\pi t\,\sin (n - \tfrac 12)\pi t\,dt = \frac 12\int _0^1\bigl [\cos (m - n)\pi t - \cos (m + n - 1)\pi t\bigr ]dt = 0 , \] since both \(m - n\) and \(m + n - 1\) are non-zero integers.

Solution. \(y = \sin kt\), and \(k\cos k + \sin k = 0\), i.e. \(\tan k = -k\). The graphs of \(\tan k\) and \(-k\) cross once in each interval \(((n - \frac 12)\pi , n\pi )\): \(k_1 = 2.0288\) and \(k_2 = 4.9132\), so \(\lambda _1 = 4.116\) and \(\lambda _2 = 24.14\). For large \(n\), \(k_n \to (n - \frac 12)\pi \) from above. (\(\lambda \leq 0\) gives no eigenvalue: for \(\lambda = -\mu ^2\), \(y = \sinh \mu t\) and \(\mu \cosh \mu + \sinh \mu > 0\).)

Solution. \(u_1 = t\) satisfies the condition at \(0\), \(u_2 = 1\) the condition at \(1\); with \(p = 1\), \(C = u_1'u_2 - u_1u_2' = 1\). So \(G(t, s) = \min (t, s)\). For \(f = 1\), \(y(t) = \int _0^1\min (t, s)\,ds = \int _0^ts\,ds + t(1 - t) = t - \frac {t^2}{2}\), which satisfies \(-y'' = 1\), \(y(0) = 0\) and \(y'(1) = 0\).

Solution. By Theorem 9.2.2, with \(\phi = \sin t\), the condition is \(\int _0^\pi (c + \cos 2t)\sin t\,dt = 0\). Now \(\int _0^\pi \sin t = 2\) and \[ \int _0^\pi \cos 2t\sin t\,dt = \int _0^\pi \tfrac 12(\sin 3t - \sin t)\,dt = \tfrac 12\bigl (\tfrac 23 - 2\bigr ) = -\tfrac 23 . \] So \(2c - \frac 23 = 0\), \(c = \frac 13\).

Solution. \(\int _0^1\pi ^2\cos ^2\pi t\,dt = \frac {\pi ^2}{2}\), and \[ \int _0^1(1 + t)\sin ^2\pi t\,dt = \frac 12 + \int _0^1t\sin ^2\pi t\,dt = \frac 12 + \frac 14 = \frac 34 . \] So \(R = \frac {\pi ^2/2}{3/4} = \frac {2\pi ^2}{3} = 6.580\), above \(\lambda _1 = 6.548\) by only \(0.5\%\), better than the polynomial trial function \(t(1 - t)\) of Software 9.6.2.

Solution. On \([a, \infty )\), \(q = 1 + t \geq 1 + a = m^2\). By Corollary 9.5.2(ii), consecutive zeros there are at most \(\pi /m = \pi /\sqrt {1 + a}\) apart, so there are infinitely many, and they become denser as \(t\) increases.