The solution of the system $\frac { d x } { d t } = - 6 x - 5 y , \frac { d y } { d t } = 4 x + 2 y$ is
A) $x = c _ { 1 } e ^ { - 2 t } ( 2 \cos ( 2 t ) + 4 \sin ( 2 t ) ) + c _ { 2 } e ^ { - 2 t } ( 4 \cos ( 2 t ) - 2 \sin ( 2 t ) )$ , $y = 4 c _ { 1 } e ^ { - 2 t } \cos ( 2 t ) + 4 c _ { 2 } e ^ { - 2 t } \sin ( 2 t )$
B) $x = c _ { 1 } e ^ { 2 t } ( 2 \cos ( 2 t ) + 4 \sin ( 2 t ) ) + c _ { 2 } e ^ { 2 t } ( 4 \cos ( 2 t ) - 2 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { 2 t } \sin ( 2 t )$
C) $x = c _ { 1 } e ^ { - 2 t } ( 4 \cos ( 2 t ) + 2 \sin ( 2 t ) ) + c _ { 2 } e ^ { - 2 t } ( - 2 \cos ( 2 t ) + 4 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { - 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { - 2 t } \sin ( 2 t )$
D) $x = c _ { 1 } e ^ { - 2 t } ( 2 \cos ( 2 t ) + 4 \sin ( 2 t ) ) + c _ { 2 } e ^ { - 2 t } ( 4 \cos ( 2 t ) - 2 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { - 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { - 2 t } \sin ( 2 t )$
E) $x = c _ { 1 } e ^ { 2 t } ( 4 \cos ( 2 t ) + 2 \sin ( 2 t ) ) + c _ { 2 } e ^ { 2 t } ( - 2 \cos ( 2 t ) + 4 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { 2 t } \sin ( 2 t )$

Question

Quizplus · Accepted Answer

The Answer of The solution of the system $\frac { d x } { d t } = - 6 x - 5 y , \frac { d y } { d t } = 4 x + 2 y$ is
A) $x = c _ { 1 } e ^ { - 2 t } ( 2 \cos ( 2 t ) + 4 \sin ( 2 t ) ) + c _ { 2 } e ^ { - 2 t } ( 4 \cos ( 2 t ) - 2 \sin ( 2 t ) )$ , $y = 4 c _ { 1 } e ^ { - 2 t } \cos ( 2 t ) + 4 c _ { 2 } e ^ { - 2 t } \sin ( 2 t )$
B) $x = c _ { 1 } e ^ { 2 t } ( 2 \cos ( 2 t ) + 4 \sin ( 2 t ) ) + c _ { 2 } e ^ { 2 t } ( 4 \cos ( 2 t ) - 2 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { 2 t } \sin ( 2 t )$
C) $x = c _ { 1 } e ^ { - 2 t } ( 4 \cos ( 2 t ) + 2 \sin ( 2 t ) ) + c _ { 2 } e ^ { - 2 t } ( - 2 \cos ( 2 t ) + 4 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { - 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { - 2 t } \sin ( 2 t )$
D) $x = c _ { 1 } e ^ { - 2 t } ( 2 \cos ( 2 t ) + 4 \sin ( 2 t ) ) + c _ { 2 } e ^ { - 2 t } ( 4 \cos ( 2 t ) - 2 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { - 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { - 2 t } \sin ( 2 t )$
E) $x = c _ { 1 } e ^ { 2 t } ( 4 \cos ( 2 t ) + 2 \sin ( 2 t ) ) + c _ { 2 } e ^ { 2 t } ( - 2 \cos ( 2 t ) + 4 \sin ( 2 t ) )$ , $y = - 4 c _ { 1 } e ^ { 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { 2 t } \sin ( 2 t )$

The Solution of the System dxdt=−6x−5y,dydt=4x+2y\frac { d x } { d t } = - 6 x - 5 y , \frac { d y } { d t } = 4 x + 2 ydtdx​=−6x−5y,dtdy​=4x+2y

The Solution of the System $\frac { d x } { d t } = - 6 x - 5 y , \frac { d y } { d t } = 4 x + 2 y$