91Ó°ÊÓ

The Wronskian of the given vectors is given by the determinant of the matrix formed by placing the vectors as columns: $$ W\left(\mathbf{x}^{(1)}, \mathbf{x}^{(2)}\right) = \begin{vmatrix} t & t^2 \\ 1 & 2t \end{vmatrix} $$ Now we can compute the Wronskian by calculating the determinant: $$ W\left(\mathbf{x}^{(1)}, \mathbf{x}^{(2)}\right) = t(2t) - (1)(t^2) = t^2 $$

The given vectors \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) are linearly independent if their Wronskian is nonzero. Since \(W(\mathbf{x}^{(1)}, \mathbf{x}^{(2)}) = t^2\), the vectors are linearly independent for all \(t \neq 0\). Thus, the intervals of linear independence are \((-\infty, 0)\) and \((0, \infty)\).

The system of homogeneous differential equations satisfied by \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) must have constant coefficients, as their linear independence depends on the interval of \(t\) (excluding \(t=0\)). This implies that the coefficients in the system of equations do not depend on \(t\).

The vectors \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) satisfy a system of first-order linear constant-coefficient homogeneous equations of the form: $$ \begin{aligned} a_1\frac{d\mathbf{x}^{(1)}(t)}{dt} + a_2\frac{d\mathbf{x}^{(2)}(t)}{dt} &= 0, \\ b_1\frac{d\mathbf{x}^{(1)}(t)}{dt} + b_2\frac{d\mathbf{x}^{(2)}(t)}{dt} &= 0 \\ \end{aligned} $$ Find the derivatives of the given vectors \({\mathbf{x}^{(1)}(t)}\) and \({\mathbf{x}^{(2)}(t)}\): $$ \frac{d\mathbf{x}^{(1)}(t)}{dt} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad \frac{d\mathbf{x}^{(2)}(t)}{dt} = \begin{pmatrix} 2t \\ 2 \end{pmatrix} $$ Now, substitute these derivatives into the system of equations. The resulting equations simplify to: $$ \begin{aligned} a_1 +2ta_2 = 0, \\ 2b_1 + 2tb_2 = 0 \\ \end{aligned} $$ Solving for \(a_1\) and \(b_1\), we find \(a_1=-2ta_2\) and \(b_1=-tb_2\). Since they both depend on \(t\), the system of differential equations is: $$ \begin{aligned} -2ta_2\frac{d\mathbf{x}^{(1)}(t)}{dt} + a_2\frac{d\mathbf{x}^{(2)}(t)}{dt} &= 0, \\ -tb_2\frac{d\mathbf{x}^{(1)}(t)}{dt} + b_2\frac{d\mathbf{x}^{(2)}(t)}{dt} &= 0 \\ \end{aligned} $$ The above system of equations confirms the conclusion from part c that the coefficients in the system of homogeneous differential equations do not depend on \(t\) (excluding the \(t=0\)).