Question

7. The general solution of the system $egin{bmatrix} x' y' end{bmatrix} = P egin{bmatrix} x y end{bmatrix} + egin{bmatrix} e^{-4t} 0 end{bmatrix}$ (where P is a 2 x 2 constant matrix) looks like $egin{bmatrix} x y end{bmatrix} = C_1 egin{bmatrix} -1 1 end{bmatrix} e^{-4t} + C_2 egin{bmatrix} 1 1 end{bmatrix} e^{-2t} + vec{V}$. If you use variation of parameter to calculate $vec{V}$, which of the following options could be $vec{V}$? (a) $frac{1}{4} egin{bmatrix} 2t - 1 -2t end{bmatrix} e^{-4t}$ (b) $frac{1}{4} egin{bmatrix} 2t -2t - 1 end{bmatrix} e^{-4t}$ (c) $frac{1}{4} egin{bmatrix} 2t - 1 -2t - 1 end{bmatrix} e^{-4t}$ (d) $frac{1}{4} egin{bmatrix} -2t - 1 2t - 1 end{bmatrix} e^{-4t}$

          7. The general solution of the system
$egin{bmatrix} x'  y' end{bmatrix} = P egin{bmatrix} x  y end{bmatrix} + egin{bmatrix} e^{-4t}  0 end{bmatrix}$
(where P is a 2 x 2 constant matrix) looks like
$egin{bmatrix} x  y end{bmatrix} = C_1 egin{bmatrix} -1  1 end{bmatrix} e^{-4t} + C_2 egin{bmatrix} 1  1 end{bmatrix} e^{-2t} + vec{V}$.
If you use variation of parameter to calculate $vec{V}$, which of the following options could
be $vec{V}$?
(a) $frac{1}{4} egin{bmatrix} 2t - 1  -2t end{bmatrix} e^{-4t}$
(b) $frac{1}{4} egin{bmatrix} 2t  -2t - 1 end{bmatrix} e^{-4t}$
(c) $frac{1}{4} egin{bmatrix} 2t - 1  -2t - 1 end{bmatrix} e^{-4t}$
(d) $frac{1}{4} egin{bmatrix} -2t - 1  2t - 1 end{bmatrix} e^{-4t}$

$7. The general solution of the system eginbmatrix x' y' endbmatrix = P eginbmatrix x y endbmatrix + eginbmatrix e^-4t 0 endbmatrix (where P is a 2 x 2 constant matrix) looks like eginbmatrix x y endbmatrix = C1 eginbmatrix -1 1 endbmatrix e^-4t + C2 eginbmatrix 1 1 endbmatrix e^-2t + vecV. If you use variation of parameter to calculate vecV, which of the following options could be vecV? (a) frac14 eginbmatrix 2t - 1 -2t endbmatrix e^-4t (b) frac14 eginbmatrix 2t -2t - 1 endbmatrix e^-4t (c) frac14 eginbmatrix 2t - 1 -2t - 1 endbmatrix e^-4t (d) frac14 eginbmatrix -2t - 1 2t - 1 endbmatrix e^-4t$

Added by David P.

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