Solution. \(t^2 \cdot 0 - 2t + 2t = 0\) and \(t^2 \cdot 2 - 2t \cdot 2t + 2t^2 = 0\). Directly, \(W = t \cdot 2t - t^2 \cdot 1 = t^2\). In normal form the equation is \(y'' - (2/t)y' + (2/t^2)y = 0\), so \(a_1 = -2/t\), and Abel gives \(W(t) = W(1)\exp \int _1^t2/s\,ds = 1 \cdot t^2\).

Solution. Their Wronskian is \(W = \sin t \cdot 1 - t\cos t\), which is \(0\) at \(t = 0\) but not identically zero (\(W'(t) = t\sin t > 0\) for \(t \neq 0\) near \(0\), so \(W \neq 0\) for small \(t \neq 0\)). By Abel’s identity the Wronskian of two solutions is either never zero or always zero, so they cannot both be solutions.

Solution. With \(x = (y, y')\), \(\Phi = \begin {bmatrix} y_1 & y_2\\ y_1' & y_2'\end {bmatrix}\) and forcing \((0, g)^{\mathsf T}\), the condition \(\Phi c' = (0, g)^{\mathsf T}\) gives \(c_1' = -y_2g/W\), \(c_2' = y_1g/W\) by Cramer’s rule, and \(y_p = c_1y_1 + c_2y_2\), which is the stated formula. For \(\tan t\), with \(y_1 = \cos t\), \(y_2 = \sin t\), \(W = 1\): \[ c_1 = -\int \sin t\tan t\,dt = \sin t - \log |\sec t + \tan t|, \qquad c_2 = \int \sin t\,dt = -\cos t , \] so \[ y_p = \cos t\sin t - \cos t\log |\sec t + \tan t| - \sin t\cos t = -\cos t\log |\sec t + \tan t| . \]

Solution. In normal form \(p = -2t/(1 - t^2)\), so \(e^{-\int p} = e^{-\log (1 - t^2)} = 1/(1 - t^2)\). Then \begin {align*} y_2 &= t\int \frac {dt}{t^2(1 - t^2)} = t\int \Bigl (\frac {1}{t^2} + \frac {1}{1 - t^2}\Bigr )dt\\ &= t\Bigl (-\frac 1t + \frac 12\log \frac {1 + t}{1 - t}\Bigr ) = \frac t2\log \frac {1 + t}{1 - t} - 1 , \end {align*}

the Legendre function of the second kind \(Q_1\), which is singular at \(t = \pm 1\).

Solution. By Theorem 4.4.1, \(y^{\mathsf T}x\) is constant, so it equals its value \(c^{\mathsf T}x(T)\) at \(t = T\). In particular \(c^{\mathsf T}x(T) = y(t_0)^{\mathsf T}x(t_0)\), so the gradient of \(c^{\mathsf T}x(T)\) with respect to \(x(t_0)\) is \(y(t_0)\). One backward integration of the adjoint from \(y(T) = c\) gives all \(n\) partial derivatives, where finite differences would need \(n + 1\) forward integrations.

Solution. \(B(t) = \bigl (\int _{t_0}^ta\bigr )A_0\) is a scalar multiple of \(A_0\), which commutes with \(A(t) = a(t)A_0\), so Proposition 4.5.1 applies. Here \(\int _1^tds/s = \log t\), and since \(A_0^2 = I\), \(e^{\tau A_0} = \cosh \tau \,I + \sinh \tau \,A_0\). With \(\tau = \log t\), \(\cosh \tau = \frac 12(t + t^{-1})\) and \(\sinh \tau = \frac 12(t - t^{-1})\), so \(\Phi (t, 1) = \frac 12\begin {bmatrix} t + t^{-1} & t - t^{-1}\\ t - t^{-1} & t + t^{-1}\end {bmatrix}\). Its determinant is \(1\), as Liouville’s formula requires since \(\tr A = 0\).