91Ó°ÊÓ

Form the matrix \(A\) by taking the coefficients of \(x_1\) and \(x_2\) in the system of differential equations. In this case, the matrix \(A\) is: $$A = \begin{pmatrix}-4 & 3 \\ 6 & -4\end{pmatrix}$$ Now, create the column vector \(\mathbf{b}(t)\) using the time-dependent terms of the given system of differential equations: $$\mathbf{b}(t) = \begin{pmatrix} 4t \\ t^2 \end{pmatrix}$$

Now, write the vector equation with the matrix \(A\) and the vector \(\mathbf{b}(t)\) like so: $$\mathbf{x}'(t) = A \cdot \mathbf{x}(t) + \mathbf{b}(t)$$, where \(\mathbf{x}'(t)\) is the time derivative of the vector \(\mathbf{x}(t)\). Putting everything together, we have: $$\begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} -4 & 3 \\ 6 & -4 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 4t \\ t^2 \end{pmatrix}$$ Thus, the vector formulation for the given system of differential equations is: $$\mathbf{x}'(t) = A \cdot \mathbf{x}(t) + \mathbf{b}(t) = \begin{pmatrix} -4 & 3 \\ 6 & -4 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 4t \\ t^2 \end{pmatrix}$$