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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Definitions:

Qualitative

Relating to, measuring, or measured by the quality of something rather than its quantity.

Categorical

Pertaining to data that can be divided into specific groups or categories that lack a natural ordering.

Nominal Variable

A type of variable that represents categories with no intrinsic ordering or ranking among them, used for classification or grouping.

Variable

An element, feature, or factor that is liable to vary or change.