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177_pdfsam_math 54 differential equation solutions odd

# 177_pdfsam_math 54 differential equation solutions odd - |...

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Exercises 4.2 31. Using the trigonometric identity 1 + tan 2 t sec 2 t , we conclude that y 1 ( t ) = tan 2 t sec 2 t ≡ − 1 y 2 ( t ) 3 ( 3) y 1 ( t ) , and so y 1 ( t ) and y 2 ( t ) are linearly dependent on (0 , 1) (even on ( −∞ , )). 33. (a) True. Since y 1 ( t ) and y 2 ( t ) are linearly dependent on [ a, b ], there exists a constant c such that y 1 ( t ) = cy 2 ( t ) (or y 2 ( t ) = cy 1 ( t )) for all t in [ a, b ]. In particular, this equality is satisfied on any smaller interval [ c, d ], and so y 1 ( t ) and y 2 ( t ) are linearly dependent on [ c, d ]. (b) False. As an example, consider y 1 ( t ) = t and y 2 ( t ) = | t | on [ 1 , 1]. For t in [0 , 1], y 2 ( t ) = t = y 1 ( t ), and so y 2 ( t ) c 1 y 1 ( t ) with constant c 1 = 1. For t in [ 1 , 0], we have y 2 ( t ) = t = y 1 ( t ), and so y 2 ( t ) c 2 y 1 ( t ) with constant c 2 = 1. Therefore, these two functions are linearly dependent on [0 , 1] and on [ 1 , 0]. Since c 1 = c 2 , there is no such a constant c that y 1 ( t ) cy 2 ( t ) on [ 1 , 1]. So, y 1 ( t ) and y 2 ( t ) are linearly independent on [ 1 , 1]. 35. (a) No, because, for t 0, y 2 ( t ) = | t 3 | = t 3 = y 1 ( t ). (b)
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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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