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Solve the given initial-value problem.\\ $x' = \begin{pmatrix} 4 & -1 \\ 5 & 2 \end{pmatrix} x$, $x(0) = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$\ $x(t) = -6 < \cos(2t), \cos(2t) + 2 \sin(2t) > e^{7t} - 3 < \sin(2t), \sin(2t) - 2 \cos(2t) > e^{7t}$

          Solve the given initial-value problem.\\
$x' = \begin{pmatrix} 4 & -1 \\ 5 & 2 \end{pmatrix} x$, $x(0) = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$\
$x(t) = -6 < \cos(2t), \cos(2t) + 2 \sin(2t) > e^{7t} - 3 < \sin(2t), \sin(2t) - 2 \cos(2t) > e^{7t}$

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Solve the given initial-value problem.

x' =
< p m a t r i x >
x, x(0) =
< p m a t r i x >x(t) = -6 < cos(2t), cos(2t) + 2 sin(2t) > e^7t - 3 < sin(2t), sin(2t) - 2 cos(2t) > e^7t

Added by Asunci-N B.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Sri K

06/09/2023

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The given initial-value problem is K' = X(0) with the initial condition X(t) = X. Show more…

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