nagle_differential_equations_ISM_Part49

# nagle_differential_equations_ISM_Part49 - Chapter 7...

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Unformatted text preview: Chapter 7 Replacing in Problem 52 of this section n by n + 1 yields L- 1 1 s n +(3 / 2) ( t ) = 2 n +1 t n +(1 / 2) 1 Â· 3 Â· 5 Â·Â·Â· (2 n + 1) âˆš Ï€ , so that L- 1 s- 3 / 2 e- 1 /s = L- 1 âˆž X n =0 (- 1) n n ! s n +3 / 2 = âˆž X n =0 (- 1) n n ! L- 1 1 s n +(3 / 2) = âˆž X n =0 (- 1) n n ! 2 n +1 t n +(1 / 2) 1 Â· 3 Â· 5 Â·Â·Â· (2 n + 1) âˆš Ï€ . Multiplying the numerator and denominator of the n th term by 2 Â· 4 Â·Â·Â· (2 n ) = 2 n n !, we obtain L- 1 s- 3 / 2 e- 1 /s ( t ) = âˆž X n =0 (- 1) n 2 n +1 2 n t n +(1 / 2) (2 n + 1)! âˆš Ï€ = âˆž X n =0 (- 1) n 2 2 n +1 (2 n + 1)! âˆš Ï€ t (2 n +1) / 2 = 1 âˆš Ï€ âˆž X n =0 (- 1) n (2 âˆš t ) 2 n +1 (2 n + 1)! = 1 âˆš Ï€ sin 2 âˆš t . (See (7.17).) 58. (a) Since u ( t- a ) = , t < a 1 , t > a, we have (i) for t < 0, u ( t )- u ( t- a ) = 0- 0 = 0; (ii) for 0 < t < a , u ( t )- u ( t- a ) = 1- 0 = 1; (iii) for t > a , u ( t )- u ( t- a ) = 1- 1 = 0 . Thus, u ( t )- u ( t- a ) = G a ( t ). (b) We use now formula (4) from the text to get L{ G a } ( s ) = L{ u ( t )- u ( t- a ) } ( s ) = L{ u ( t ) } ( s )- L{ u ( t- a ) } ( s ) = 1 s- e- as s = 1- e- as s . 236 Exercises 7.6 (c) Since G a ( t- b ) = u ( t- b )- u [( t- b )- a ] = u ( t- b )- u [ t- ( a + b )] , similarly to part (b) we have L{ G a ( t- b ) } ( s ) = L{ u ( t- b )- u [ t- ( a + b )] } ( s ) = L{ u ( t- b ) } ( s )- L{ u [ t- ( a + b )] } ( s ) = e- bs s- e- ( a + b ) s s = e- bs- e- ( a + b ) s s ....
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nagle_differential_equations_ISM_Part49 - Chapter 7...

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