Question

x'(t) = \begin{bmatrix} -9 & -1 \\ 10 & -7 \end{bmatrix} x(t), \\ (a) x(0) = \begin{bmatrix} -12 \\ 0 \end{bmatrix} \qquad (b) x(\pi) = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \qquad (c) x(-2\pi) = \begin{bmatrix} 4 \\ 1 \end{bmatrix} \qquad (d) x(\pi/6) = \begin{bmatrix} 0 \\ 3 \end{bmatrix} \\ (a) x(t) = \boxed{} \\ (Use parentheses to clearly denote the argument of each function.)

          x'(t) = \begin{bmatrix} -9 & -1 \\ 10 & -7 \end{bmatrix} x(t), \\
(a) x(0) = \begin{bmatrix} -12 \\ 0 \end{bmatrix} \qquad (b) x(\pi) = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \qquad (c) x(-2\pi) = \begin{bmatrix} 4 \\ 1 \end{bmatrix} \qquad (d) x(\pi/6) = \begin{bmatrix} 0 \\ 3 \end{bmatrix} \\
(a) x(t) = \boxed{} \\
(Use parentheses to clearly denote the argument of each function.)
        
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x'(t) = 
    < b m a t r i x >
 x(t), 

(a) x(0) = 
    < b m a t r i x >
     (b) x(π) = 
    < b m a t r i x >
     (c) x(-2π) = 
    < b m a t r i x >
     (d) x(π/6) = 
    < b m a t r i x >
 

(a) x(t) =  

(Use parentheses to clearly denote the argument of each function.)

Added by William H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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x(t) = [:-1] 0 - [] [] - [] [ ax(t) = Use parentheses to clearly denote the argument of each function
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Transcript

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00:01 Okay, here we have this relationship, y times natural lock x plus xe to the y equals 1.
00:10 We want to know dx dt.
00:13 So we want to know d .ydt when dxdt is 5, given that dxdt is 5 when x is 1 and y is 0.
00:24 So let's differentiate.
00:26 So we have dydt.
00:27 Everything's being differentiated with respect to t so then y times one over x dx dt product rule another product rule dx dt e to the y plus x e to the y d y dt and then derivative 1 to 0 so i just plug in what we know d, y, dt, natural log of x, that's natural log of 1, which is zero.
01:08 It's nice...
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