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Consider the First-Order Nonhomogeneous Initial Value Problem

Given a Fundamental

Question 82

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Consider the first-order nonhomogeneous initial value problem
 Consider the first-order nonhomogeneous initial value problem   Given a fundamental matrix   (t)  for the system, what is the solution of this initial value problem? A)    \mathbf{x}(t) =\psi(3.0) \left(\begin{array}{l}-2 \\ 4\end{array}\right) +\psi(t)  \int_{3.0}^{t} \psi^{-1}(s) { }_{-5 e^{-2 s}}^{4 s^{3}+4} d s   B)    \mathbf{x}(t) =\psi(t)  \psi^{-1}(3.0) \left(\begin{array}{c}-2 \\ 4\end{array}\right) +\psi(t)  \int_{3.0}^{t} \psi^{-1}(s) { }_{-5 e^{-2 s}}^{4 s^{3}+4} d s   C)    \mathbf{x}(t) =\psi(t) \left(\begin{array}{l}-2 \\ 4\end{array}\right) +\psi(t)  \int_{3.0}^{t} \psi^{-1}(s)  \begin{array}{l}4 s^{3}+4 \\ -5 e^{-2 s}\end{array} s   D)    \mathbf{x}(t) =\psi(t)  \psi^{-1}(3.0) \left(\begin{array}{c}-2 \\ 4\end{array}\right) +\psi(3.0)  \int_{3.0}^{t} \psi^{-1}(s)  s^{3}+4 e^{-2 s} d s
Given a fundamental matrix  Consider the first-order nonhomogeneous initial value problem   Given a fundamental matrix   (t)  for the system, what is the solution of this initial value problem? A)    \mathbf{x}(t) =\psi(3.0) \left(\begin{array}{l}-2 \\ 4\end{array}\right) +\psi(t)  \int_{3.0}^{t} \psi^{-1}(s) { }_{-5 e^{-2 s}}^{4 s^{3}+4} d s   B)    \mathbf{x}(t) =\psi(t)  \psi^{-1}(3.0) \left(\begin{array}{c}-2 \\ 4\end{array}\right) +\psi(t)  \int_{3.0}^{t} \psi^{-1}(s) { }_{-5 e^{-2 s}}^{4 s^{3}+4} d s   C)    \mathbf{x}(t) =\psi(t) \left(\begin{array}{l}-2 \\ 4\end{array}\right) +\psi(t)  \int_{3.0}^{t} \psi^{-1}(s)  \begin{array}{l}4 s^{3}+4 \\ -5 e^{-2 s}\end{array} s   D)    \mathbf{x}(t) =\psi(t)  \psi^{-1}(3.0) \left(\begin{array}{c}-2 \\ 4\end{array}\right) +\psi(3.0)  \int_{3.0}^{t} \psi^{-1}(s)  s^{3}+4 e^{-2 s} d s (t) for the system, what is the solution of this initial value problem?


A) x(t) =ψ(3.0) (24) +ψ(t) 3.0tψ1(s) 5e2s4s3+4ds \mathbf{x}(t) =\psi(3.0) \left(\begin{array}{l}-2 \\ 4\end{array}\right) +\psi(t) \int_{3.0}^{t} \psi^{-1}(s) { }_{-5 e^{-2 s}}^{4 s^{3}+4} d s
B) x(t) =ψ(t) ψ1(3.0) (24) +ψ(t) 3.0tψ1(s) 5e2s4s3+4ds \mathbf{x}(t) =\psi(t) \psi^{-1}(3.0) \left(\begin{array}{c}-2 \\ 4\end{array}\right) +\psi(t) \int_{3.0}^{t} \psi^{-1}(s) { }_{-5 e^{-2 s}}^{4 s^{3}+4} d s
C) x(t) =ψ(t) (24) +ψ(t) 3.0tψ1(s) 4s3+45e2ss \mathbf{x}(t) =\psi(t) \left(\begin{array}{l}-2 \\ 4\end{array}\right) +\psi(t) \int_{3.0}^{t} \psi^{-1}(s) \begin{array}{l}4 s^{3}+4 \\ -5 e^{-2 s}\end{array} s
D) x(t) =ψ(t) ψ1(3.0) (24) +ψ(3.0) 3.0tψ1(s) s3+4e2sds \mathbf{x}(t) =\psi(t) \psi^{-1}(3.0) \left(\begin{array}{c}-2 \\ 4\end{array}\right) +\psi(3.0) \int_{3.0}^{t} \psi^{-1}(s) s^{3}+4 e^{-2 s} d s

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