Chapter 5: Problem 3
Show that linearization is a well-defined operation: the operator \(A\) does not depend on the coordinate system. The advantage of the linearized system is that it is linear and therefore easily solved: \(\mathbf{y}(t)=e^{A t} \mathbf{y}(0), \quad\) where \(e^{A t}=E+A t+\frac{A^{2} t^{2}}{2 !}+\cdots .\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.