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Pre-Calc Homework Solutions 238

Pre-Calc Homework Solutions 238 - 238 Section 6.1 d 2 x(x e...

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49. continued (b) Again, y 5 } x 2 2 } 1 } 1 x } 1 C . Initial condition: y ( 2 1) 5 1 1 5 } ( 2 2 1) 2 } 1 } ( 2 1 1) } 1 C 1 5 2 } 1 2 } 1 C } 3 2 } 5 C Solution: y 5 } x 2 2 } 1 } 1 x } 1 } 3 2 } , x , 0 (c) For x , 0, } d d y x } 5 } d d x } 1 } 1 x } 1 } x 2 2 } 1 C 1 2 5 2 } x 1 2 } 1 x 5 x 2 } x 1 2 } . For x . 0, } d d y x } 5 } d d x } 1 } 1 x } 1 } x 2 2 } 1 C 2 2 5 2 } x 1 2 } 1 x 5 x 2 } x 1 2 } . And for x 5 0, } d d y x } is undefined. (d) Let C 1 be the value from part (b), and let C 2 be the value from part (a). Thus, C 1 5 } 3 2 } and C 2 5 } 1 2 } . (e) y (2) 5 2 1 y ( 2 2) 5 2 2 1 5 } 1 2 } 1 } 2 2 2 } 1 C 2 2 5 } ( 2 1 2) } 1 } ( 2 2 2) 2 } 1 C 1 2 1 5 } 5 2 } 1 C 2 2 5 } 3 2 } 1 C 1 2 } 7 2 } 5 C 2 } 1 2 } 5 C 1 Thus, C 1 5 } 1 2 } and C 2 5 2 } 7 2 } . 50. E } d d x r } dx 5 E (3 x 2 2 6 x 1 12) dx r 5 x 3 2 3 x 2 1 12 x 1 C Initial condition: r (0) 5 0 0 5 0 3 2 3(0) 2 1 12(0) 1 C 0 5 C Solution: r ( x ) 5 x 3 2 3 x 2 1 12 x 51. E } d d c x } dx 5 E (3 x 2 2 12 x 1 15) dx c 5 x 3 2 6 x 2 1 15 x 1 C Initial condition c (0) 5 400 400 5 0 3 2 6(0) 2 1 15(0) 1 C 400 5 C Solution: c ( x ) 5 x 3 2 6 x 2 1 15 x 1 400 52. (a) E f ( x ) dx 5 E } d d x } ( x 2 e x ) dx 5 x 2 e x 1 C (b) E g ( x ) dx 5 E } d d x } ( x sin x ) dx 5 x sin x 1 C (c) E [ 2 f ( x )] dx 5 2 E f ( x ) dx 5 2 x 2 e x 1 C (d) E [ 2 g ( x )] dx 5 2 E g ( x ) dx 5 2 x sin x 1 C (e) E [ f ( x ) 1 g ( x )] dx 5 E f ( x ) dx 1 E g ( x ) dx 5 x 2 e x 1 x sin x 1 C (f) E [ f ( x ) 2 g ( x )] dx 5 E f ( x ) dx 2 E g ( x ) dx 5 x 2 e x 2 x sin x 1 C (g) E [ x 1 f ( x )] dx 5 E x dx 1 E f ( x ) dx 5 } x 2 2 } 1 x 2 e x 1 C (h) E [ g ( x ) 2 4] dx 5 E g ( x ) dx 2 E 4 dx 5 x sin x 2 4 x 1 C 53. (a) E } d dt 2 2 s } dt 5 E 2 k dt } d d s t } 5 2 kt 1 C 1 Initial condition: } d d s t } 5 88 when
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