$$\boldsymbol{\vec{y}}(t) = c_1 e^{2t} \left\lbrack \begin{array}{r} 1 \\ 1 \end{array} \right\rbrack + c_2 e^{t} \left\lbrack \begin{array}{r} 0 \\ 1 \end{array} \right\rbrack,$$

$$\boldsymbol{\vec{y}}(0) = \left\lbrack \begin{array}{r} -3 \\ 2 \end{array} \right\rbrack.$$
(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) Is the solution curve headed toward or away from the origin as $t$ increases?

A. toward
B. away
C. neither toward nor away