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Pre-Calc Homework Solutions 350

Pre-Calc Homework Solutions 350 - 350 Section 8.4 ds 2s ds...

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30. } 2 s d 1 s 2 } 5 } t 2 1 dt 2 t } E } 2 s d 1 s 2 } 5 } 1 2 } E } s 1 ds 1 } 5 } 1 2 } ln ) s 1 1 ) 1 C 1 t 2 1 2 t 5 t ( t 1 2) } t 2 1 1 2 t } 5 } A t } 1 } t 1 B 2 } 1 5 A ( t 1 2) 1 Bt 1 5 ( A 1 B ) t 1 2 A Equating coefficients of like terms gives A 1 B 5 0 and 2 A 5 1 Solving the system simultaneously yields A 5 } 1 2 } , B 5 2 } 1 2 } . E } t 2 1 dt 2 t } 5 E } 1/ t 2 } dt 2 E } t 1 1 /2 2 } dt 5 } 1 2 } ln ) t ) 2 } 1 2 } ln ) t 1 2 ) 1 C 2 } 1 2 } ln ) s 1 1 ) 5 } 1 2 } ln ) t ) 2 } 1 2 } ln ) t 1 2 ) 1 C 3 ln ) s 1 1 ) 5 ln ) t ) 2 ln ) t 1 2 ) 1 C Substitute t 5 1, s 5 1. ln 2 5 0 2 ln 3 1 C or C 5 ln 2 1 ln 3 5 ln 6 The solution to the initial value problem is ln ) s 1 1 ) 5 ln ) t ) 2 ln ) t 1 2 ) 1 ln 6 ln ) s 1 1 ) 5 ln ) } t 1 6 t 2 } ) ) s 1 1 ) 5 ) } t 1 6 t 2 } ) . 31. (a) Complete the square in the denominator. 5 1 4 x 2 x 2 5 5 2 ( x 2 2 4 x ) 5 9 2 ( x 2 2 4 x 1 4) 5 9 2 ( x 2 2) 2 Let u 5 x 2 2 so du 5 dx , and then use Formula 18 with x 5 u and a 5 3. E } 5 1 4 d x x 2 x 2 } 5 E } 9 2 du u 2 } 5 } 1 6 } ln ) } u u 1 2 3 3 } ) 1 C 5 } 1 6 } ln ) } x x 1 2 1 5 } ) 1 C (b) } d d x } 1 } 2 1 a } ln ) } x x 1 2 a a } ) 1 C 2 5 } 2 1 a } } d d x } 1 ln ) x 1 a ) 2 ln ) x 2 a ) 2 5 } 2 1 a } 1 } x 1 1 a } 2 } x 2 1 a } 2 5 } 2 1
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