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Consider the First-Order Nonhomogeneous System of Linear Differential Equations
x(t)=ψ(0)x0+ψ(t)0tψ1(s)g(s)ds x(t)=\psi(0) \mathbf{x}_{0}+\psi(t) \int_{0}^{t} \psi^{-1}(s) \mathbf{g}(s) d s

Question 86

Multiple Choice

Consider the first-order nonhomogeneous system of linear differential equations
 Consider the first-order nonhomogeneous system of linear differential equations   Where the components of g(t)  are continuous functions.Given a fundamental matrix   (t)  for the system, what is the solution of this system if it is equipped with the initial condition x(3.6)  = X <sub>0</sub>? A)    x(t) =\psi(0)  \mathbf{x}_{0}+\psi(t)  \int_{0}^{t} \psi^{-1}(s)  \mathbf{g}(s)  d s   B)    x(t) =\psi(t)  \psi^{-1}(3.6)  \mathbf{x}_{0}+\psi(t)  \int_{3.6}^{t} \psi^{-1}(s)  g(s)  d s   C)    x(t) =\psi(3.6)  \mathbf{x}_{0}+\psi(t)  \int_{3.6}^{t} \psi^{-1}(s)  \mathbf{g}(s)  d s   D)    x(t) =\psi(t)  \psi^{-1}(0)  \mathbf{x}_{0}+\psi(t)  \int_{0}^{t} \psi^{-1}(s)  g(s)  d s
Where the components of g(t) are continuous functions.Given a fundamental matrix  Consider the first-order nonhomogeneous system of linear differential equations   Where the components of g(t)  are continuous functions.Given a fundamental matrix   (t)  for the system, what is the solution of this system if it is equipped with the initial condition x(3.6)  = X <sub>0</sub>? A)    x(t) =\psi(0)  \mathbf{x}_{0}+\psi(t)  \int_{0}^{t} \psi^{-1}(s)  \mathbf{g}(s)  d s   B)    x(t) =\psi(t)  \psi^{-1}(3.6)  \mathbf{x}_{0}+\psi(t)  \int_{3.6}^{t} \psi^{-1}(s)  g(s)  d s   C)    x(t) =\psi(3.6)  \mathbf{x}_{0}+\psi(t)  \int_{3.6}^{t} \psi^{-1}(s)  \mathbf{g}(s)  d s   D)    x(t) =\psi(t)  \psi^{-1}(0)  \mathbf{x}_{0}+\psi(t)  \int_{0}^{t} \psi^{-1}(s)  g(s)  d s (t) for the system, what is the solution of this system if it is equipped with the initial condition x(3.6) = X 0?


A) x(t) =ψ(0) x0+ψ(t) 0tψ1(s) g(s) ds x(t) =\psi(0) \mathbf{x}_{0}+\psi(t) \int_{0}^{t} \psi^{-1}(s) \mathbf{g}(s) d s
B) x(t) =ψ(t) ψ1(3.6) x0+ψ(t) 3.6tψ1(s) g(s) ds x(t) =\psi(t) \psi^{-1}(3.6) \mathbf{x}_{0}+\psi(t) \int_{3.6}^{t} \psi^{-1}(s) g(s) d s
C) x(t) =ψ(3.6) x0+ψ(t) 3.6tψ1(s) g(s) ds x(t) =\psi(3.6) \mathbf{x}_{0}+\psi(t) \int_{3.6}^{t} \psi^{-1}(s) \mathbf{g}(s) d s
D) x(t) =ψ(t) ψ1(0) x0+ψ(t) 0tψ1(s) g(s) ds x(t) =\psi(t) \psi^{-1}(0) \mathbf{x}_{0}+\psi(t) \int_{0}^{t} \psi^{-1}(s) g(s) d s

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