Consider the first-order homogeneous system of linear differential equations

Given a fundamental matrix (t) for the system, which of these is the general solution of this system?
Here, is an arbitrary constant vector.
A) $ \mathbf{x}(t)=\psi^{-1}(t) \mathbf{C} $
B) $ x(t)=\psi(t) \mathbf{C} $
C) $ \mathbf{x}(t)=\psi(0) \mathbf{C} $
D) $ \mathbf{x}(t)=\psi(t)+C $

Question

Consider the first-order homogeneous system of linear differential equations
  
Given a fundamental matrix  (t) for the system, which of these is the general solution of this system?
Here,   is an arbitrary constant vector.
A) $ \mathbf{x}(t)=\psi^{-1}(t) \mathbf{C} $ 
B) $ x(t)=\psi(t) \mathbf{C} $ 
C) $ \mathbf{x}(t)=\psi(0) \mathbf{C} $ 
D) $ \mathbf{x}(t)=\psi(t)+C $

Quizplus · Accepted Answer

The Answer of Consider the first-order homogeneous system of linear differential equations
  
Given a fundamental matrix  (t) for the system, which of these is the general solution of this system?
Here,   is an arbitrary constant vector.
A) $ \mathbf{x}(t)=\psi^{-1}(t) \mathbf{C} $ 
B) $ x(t)=\psi(t) \mathbf{C} $ 
C) $ \mathbf{x}(t)=\psi(0) \mathbf{C} $ 
D) $ \mathbf{x}(t)=\psi(t)+C $

Consider the First-Order Homogeneous System of Linear Differential Equations x(t)=ψ−1(t)C \mathbf{x}(t)=\psi^{-1}(t) \mathbf{C} x(t)=ψ−1(t)C

Consider the First-Order Homogeneous System of Linear Differential Equations
$\mathbf{x}(t)=\psi^{-1}(t) \mathbf{C}$