Solution. Here \(p = \tan t\), \(P(t) = \int _0^t\tan s\,ds = -\log \cos t\), and \(e^{P} = \sec t\). By Theorem 1.2.1, \(y = \cos t\,\bigl (2 + \int _0^t\sec ^2s\,ds\bigr ) = \cos t(2 + \tan t) = 2\cos t + \sin t\). The theorem guarantees the solution on the largest interval containing \(0\) on which \(\tan t\) and \(\sec t\) are continuous, which is \((-\pi /2, \pi /2)\). The formula \(2\cos t + \sin t\) happens to extend smoothly beyond, but the equation itself is undefined at \(\pm \pi /2\), so the theorem says nothing there.

Solution. For \(y_0 \neq 0\), separation gives \(-1/y = t^2/2 - 1/y_0\), so \(y = 2y_0/(2 - y_0t^2)\); for \(y_0 = 0\), \(y \equiv 0\). With \(y_0 = 1\), \(y = 2/(2 - t^2)\), which exists only on \((-\sqrt 2, \sqrt 2)\) and tends to \(+\infty \) at the ends: the solution blows up in finite time although \(f(t, y) = ty^2\) is smooth everywhere. With \(y_0 = -1\), \(y = -2/(2 + t^2)\), which exists for all \(t\) and tends to \(0\). The interval of existence of a non-linear equation depends on the initial value, unlike the linear case.

Solution. \(M_y = \cos x = N_x\). Integrating \(M\) in \(x\): \(F = x^3 + y\sin x + \psi (y)\), and \(F_y = \sin x + \psi ' = \sin x - 4y^3\) gives \(\psi = -y^4\). The solutions satisfy \(x^3 + y\sin x - y^4 = C\).

Solution. \(M_y = 2y\), \(N_x = y\), and \((M_y - N_x)/N = y/(xy) = 1/x\), a function of \(x\) alone, so \(\mu = x\) by Theorem 1.4.3. Then \((x^3 + xy^2 + x^2)\,dx + x^2y\,dy = 0\) is exact, with \(F = \frac 14x^4 + \frac 12x^2y^2 + \frac 13x^3\), and the solutions satisfy \(3x^4 + 6x^2y^2 + 4x^3 = C\).

Solution. Here \(q_0 = 1 + t^2\), \(q_1 = 0\), \(q_2 = -1\). By Theorem 1.5.2(i), \(y = t + 1/v\) with \(v' = -(0 + 2(-1)t)v - (-1) = 2tv + 1\). The integrating factor is \(e^{-t^2}\), so \(v = e^{t^2}\bigl (C + \int _0^te^{-s^2}\,ds\bigr )\) and \[ y = t + \frac {e^{-t^2}}{C + \int _0^te^{-s^2}\,ds} . \] The integral is \(\frac {\sqrt \pi }{2}\operatorname {erf}(t)\), a non-elementary function, as Theorem 1.5.2(ii) leads one to expect.

Solution. The equilibria are \(0\) and \(1\). The sign of \(f\) is positive for \(y < 0\) and for \(0 < y < 1\), and negative for \(y > 1\). So \(1\) is asymptotically stable (\(f'(1) = -1\)), and \(0\) is semi-stable (\(f'(0) = 0\)): solutions approach it from below and leave it above. Starting at \(-0.1\), the solution increases towards \(0\); at \(0.1\) it increases towards \(1\); at \(1.5\) it decreases towards \(1\). The first two show that a solution starting near \(0\) on one side tends to \(0\) and on the other side goes away from it.

Solution. The equilibria are \(0\) and \(y^* = 1 - E\) (for \(E < 1\)). With \(f'(y) = 1 - E - 2y\), \(f'(0) = 1 - E > 0\) and \(f'(1 - E) = -(1 - E) < 0\), so \(y^*\) is stable and \(0\) unstable. The yield is \(E(1 - E)\), maximised at \(E = \frac 12\) with yield \(\frac 14\), the same maximum as before. But now, for every \(E < 1\), any positive stock recovers to \(y^* > 0\): there is no cliff, because the catch falls automatically when the stock falls. Here the equilibria \(0\) and \(1 - E\) exchange stability at \(E = 1\) instead of disappearing (a transcritical bifurcation), and a fixed quota \(h\) near \(\frac 14\) is dangerous because a small error or shock pushes the stock below the unstable \(y_-\).