Unformatted text preview:  t 3  = − t 3 = − y 1 ( t ). (c) Yes, because there is no constant c such that y 2 ( t ) = cy 1 ( t ) is satisFed for all t (for positive t we have c = 1, and c = − 1 for negative t ). (d) While y 1 ( t ) = 3 t 2 on ( −∞ , ∞ ), for the derivative of y 2 ( t ) we consider three di²erent cases: t < 0, t = 0, and t > 0. ±or t < 0, y 2 ( t ) = − t 3 , y 2 ( t ) = − 3 t 2 , and so W [ y 1 , y 2 ]( t ) = ± ± ± ± ± t 3 − t 3 3 t 2 − 3 t 2 ± ± ± ± ± = t 3 ( − 3 t 2 ) − 3 t 2 ( − t 3 ) = 0 . Similarly, for t > 0, y 2 ( t ) = t 3 , y 2 ( t ) = 3 t 2 , and W [ y 1 , y 2 ]( t ) = ± ± ± ± ± t 3 t 3 3 t 2 3 t 2 ± ± ± ± ± = t 3 · 3 t 2 − 3 t 2 · t 3 = 0 . 173...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, Linear Independence, Vector Space, y1, c1 y1

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