Chapter 9 - Systems of Differential Equations - 9.1 First-Order Linear Systems - Problems - Page 587: 5

$x_2=-c_1\sin 2t+c_2\cos 2t$

$x_1'=2x_2$ $x_2'=-2x_1$ Take $x_1''=2x_2'$ Since $x_2'=-2x_1$ we can notice that $x_1''=-4x_1$ This constant coefficient differential equation has auxiliary polynomial $P(r)=r^2+4$ Consequently, $x_1(t)=c_1\cos 2t+c_2\sin 2t$ Since $x_2'=-2x_1$ $x_2=-c_1\sin 2t+c_2\cos 2t$