91Ó°ÊÓ

Firstly, let's recall Problem 15(c) in Section 7.7, which states that the solution of a nonhomogeneous linear system $$ x'(t) = Ax(t) + g(t) $$ where \(A\) is an \(n \times n\) constant matrix, and \(g: \mathbb{R} \rightarrow \mathbb{R}^n\), with given initial condition \(x(0) = x^0\), is given by $$ x(t) = \Phi(t)x^0 + \int_{0}^{t}\Phi(t-s)g(s)ds $$ where \(\Phi(t)\) is the state transition matrix, satisfying \(\Phi(0) = I\), and \(\Phi'(t) = A\Phi(t)\). The task in part (a) is to show that the given initial value problem $$ x'(t) = Ax(t) + g(t), \quad x(0) = x^0 $$ also has the solution $$ x(t) = \Phi(t)x^0 + \int_{0}^{t}\Phi(t-s)g(s)ds. $$ Since the given initial value problem and the problem from Section 7.7 have the same structure, and are both nonhomogeneous linear systems, it follows that the result from Problem 15(c) can be used directly to show the required result. Therefore, the solution to the given initial value problem is indeed $$ x(t) = \Phi(t)x^0 + \int_{0}^{t}\Phi(t-s)g(s)ds. $$

Now, let's show that $$ x(t) = e^{At}x^0 + \int_{0}^{t}e^{A(t-s)}g(s)ds. $$ The matrix exponential \(e^{At}\) has the property that its derivative is \(Ae^{At}\) and it satisfies the initial condition \(e^{A(0)} = I\), where \(I\) is the identity matrix. It can be shown that the matrix exponential \(e^{At}\) is precisely the state transition matrix \(\Phi(t)\) for the given linear system. Therefore, substituting \(\Phi(t)\) with \(e^{At}\) in the solution formula from part (a) gives us $$ x(t) = e^{At}x^0 + \int_{0}^{t}e^{A(t-s)}g(s)ds. $$

Problem 27 in Section 3.7 discusses a scalar linear non-homogeneous ODE. The general solution for such problem is given by $$ x(t) = e^{\lambda t}c + \int_{0}^{t}e^{\lambda(t-s)}g(s)ds, $$ where \(\lambda\) is a scalar constant and the function \(g(t)\) is the non-homogeneous part of the equation. Comparing this result with the result of part (b), we can observe that the matrix exponential \(e^{At}\) and its integral version play analogous roles to the scalar exponential \(e^{\lambda t}\) and its integral version in the scalar case. Both solutions have a homogeneous part (involving the exponential) and a non-homogeneous part (involving the integral), showcasing the similarity between the scalar and matrix versions of the solutions.