Solution. By Theorem 3.3.1, \(T = \int _{x_0}^\infty x^{-p}\,dx = x_0^{1-p}/(p - 1)\). Separating, \(x^{1-p}/(1 - p) = t + x_0^{1-p}/(1 - p)\), so \(x(t) = \bigl (x_0^{1-p} - (p - 1)t\bigr )^{-1/(p-1)}\), which tends to infinity as \(t \to T\). For \(x' = e^x\), \(T = \int _{x_0}^\infty e^{-x}\,dx = e^{-x_0}\), and \(x = -\log (e^{-x_0} - t)\).

Solution. \(\int _{x_0}^\infty \frac {dx}{x\log x} = \bigl [\log \log x\bigr ]_{x_0}^\infty = \infty \), so there is no blow-up, by Theorem 3.3.1. Separating, \(\log \log x = t + \log \log x_0\), so \(x(t) = x_0^{e^t}\), a double exponential: fast growth, but finite for every \(t\).

Solution. \(|f(t, x)| \leq |t| + 1\), which has the form \(a(t) + b(t)|x|\) with \(a = |t| + 1\) and \(b = 0\), and \(f\) is \(C^1\), so Theorem 3.2.1 gives existence on \(\mathbb R\). Indeed \(|x(t) - 5| \leq \int _0^{|t|}(s + 1)\,ds\).

Solution. \(u \equiv 0\) satisfies \(u' = 0 \leq t = f(t, 0)\) for \(t \geq 0\), so it is a subsolution with \(u(0) = x(0)\), and \(x \geq 0\) by Theorem 3.4.1. For \(v = \sqrt t\), \(v' = \frac {1}{2\sqrt t} \geq 0 = t - v^2 = f(t, v)\), so \(v\) is a supersolution with \(v(0) = 0\); applying the theorem on \([\delta , t_1]\) for small \(\delta > 0\), where \(x(\delta ) \leq \sqrt \delta \) (since \(x(\delta ) = \delta ^2/2 + O(\delta ^4)\)), and letting \(\delta \to 0\) gives \(x \leq \sqrt t\). A solution trapped between \(0\) and \(\sqrt t\) cannot blow up, so by Theorem 3.1.3 it exists for all \(t \geq 0\).

Solution. \(D_xf = -1 + 2x\). At \(x \equiv 0\), \(Y' = -Y\), so \(Y = e^{-t}\): perturbations of \(0\) shrink, and \(0\) is asymptotically stable. At \(x \equiv 1\), \(Y' = Y\), so \(Y = e^t\): perturbations of \(1\) grow, and \(1\) is unstable. This is the stability test of Theorem 1.6.1(iv) seen through the variational equation.

Solution. \(D_xf = \lambda \) and \(D_\lambda f = x = e^{\lambda t}\), so \(Z' = \lambda Z + e^{\lambda t}\), \(Z(0) = 0\). By Theorem 1.2.1, \(Z = e^{\lambda t}\int _0^te^{-\lambda s}e^{\lambda s}\,ds = te^{\lambda t}\), which is \(\frac {\partial }{\partial \lambda }e^{\lambda t}\).

Solution. \(x\) is a subsolution of \(y' = y^2\), which gives the upper bound \(x \leq 1/(1 - t)\) on \([0, 1)\), but no lower bound. On \([0, 1]\), \(x' \geq x^2 - 1\), so \(x\) is a supersolution of \(y' = y^2 - 1\). Its solution from \(y(0) = c > 1\) is \(y = \frac {1 + ke^{2t}}{1 - ke^{2t}}\) with \(k = \frac {c - 1}{c + 1}\), which blows up when \(e^{2t} = 1/k\), that is at \(t = \frac 12\log \frac {c + 1}{c - 1}\). This is less than \(1\) when \(\frac {c + 1}{c - 1} < e^2\), that is \(c > \frac {e^2 + 1}{e^2 - 1} = 1.313\). So for \(x(0) > 1.313\) the solution blows up before \(t = 1\). For \(x(0) = 1\) this argument is inconclusive.