$$\boldsymbol{\vec{y}}(t) = c_1 e^{-t} \left\lbrack \begin{array}{r} 1 \\ -1 \end{array} \right\rbrack + c_2 e^{t} \left\lbrack \begin{array}{r} 1 \\ 1 \end{array} \right\rbrack,$$

$$\boldsymbol{\vec{y}}(1) = \left\lbrack \begin{array}{r} -2 \\ 0 \end{array} \right\rbrack.$$
(a) Find $c_1$ and $c_2$.

$c_1 =$
$c_2 =$

(b) Sketch the phase plane trajectory that satisfies the given initial condition. Which graph most closely resembles the graph you drew? Choose A B C D

(c) What is the approximate direction of travel for the solution curve, as $t$ increases from $-\infty$ to $+\infty$?

A. along the line $y = x$ toward the origin and then along the line $y = -x$ away from the origin

B. along the line $y = -x$ toward the origin and then along the line $y = x$ away from the origin

C. none of the above