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

question 5

Multiple Choice

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) \(x = c _ { 1 } e ^ { 2 t } ( 4 \cos ( 2 t ) + 2 \sin ( 2 t ) ) + c _ { 2 } e ^ { 2 t } ( - 2 \cos ( 2 t ) + 4 \sin ( 2 t ) )\) , \(y = - 4 c _ { 1 } e ^ { 2 t } \cos ( 2 t ) - 4 c _ { 2 } e ^ { 2 t } \sin ( 2 t )\)

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