Unformatted text preview: 8 1 / 32 2 ± ± ± ± ± ± ± ± = ± ± ± ± ± 32 1 / 2 1 / 32 2 ± ± ± ± ± − ± ± ± ± ± 8 1 / 2 1 / 8 2 ± ± ± ± ± + ± ± ± ± ± 8 3 2 1 / 8 1 / 32 ± ± ± ± ± = 2827 64 6 = 0 . Hence it has the unique trivial solution, that is, c 1 = c 2 = c 3 = 0. 9. Let y 1 = sin 2 x , y 2 = cos 2 x , and y 3 = 1. We want to fnd c 1 , c 2 , and c 3 , not all zero, such that c 1 y 1 + c 2 y 2 + c 3 y 3 = c 1 sin 2 x + c 2 cos 2 x + c 3 · 1 = 0 , ±or all x in the interval ( −∞ , ∞ ). Since sin 2 x + cos 2 x = 1 ±or all real numbers x , we can choose c 1 = 1, c 2 = 1, and c 3 = − 1. Thus, these ±unctions are linearly dependent. 342...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Vector Space, basis, c1 + c2

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