nagle_differential_equations_ISM_Part44

nagle_differential_equations_ISM_Part44 - Exercises 7.4...

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Unformatted text preview: Exercises 7.4 Multiplying (7.5) by s 3 and substituting s = 0 yields 7 s 3- 2 s 2- 3 s + 6 s- 2 s =0 =- 3 = A + Bs + Cs 2 + Ds 3 s- 2 s =0 = A. Thus, A =- 3. Multiplying (7.5) by s- 2 and evaluating the result at s- 2, we get 7 s 3- 2 s 2- 3 s + 6 s 3 s =2 = 6 = ( s- 2)- 3 s 3 + B s 2 + C s + D s =2 = D . So, D = 6 and (7.5) becomes 7 s 3- 2 s 2- 3 s + 6 s 3 ( s- 2) =- 3 s 3 + B s 2 + C s + 6 s- 2 . Clearing the fractions yields 7 s 3- 2 s 2- 3 s + 6 =- 3( s- 2) + Bs ( s- 2) + Cs 2 ( s- 2) + 6 s 3 . Matching the coefficients at s 3 , we obtain C + 6 = 7 or C = 1. Finally, the coefficients at s 2 lead to B- 2 C =- 2 or B = 0. Therefore, F ( s ) = 7 s 3- 2 s 2- 3 s + 6 s 3 ( s- 2) =- 3 s 3 + 1 s + 6 s- 2 and L- 1 { F ( s ) } ( t ) =- 3 2 t 2 + 1 + 6 e 2 t . 28. First, we find F ( s ). F ( s ) ( s 2 + s- 6 ) = s 2 + 4 s 2 + s F ( s ) = s 2 + 4 s ( s + 1)( s 2 + s- 6) = s 2 + 4 s ( s + 1)( s + 3)( s- 2) . The partial fractions expansion yields s 2 + 4 s ( s + 1)( s + 3)( s- 2) = A s + B s + 1 + C s + 3 + D s- 2 . Clearing fractions gives us s 2 + 4 = A ( s + 1)( s + 3)( s- 2) + Bs ( s + 3)( s- 2) + Cs ( s + 1)( s- 2) + Ds ( s + 1)( s + 3) . With s = 0, s =- 1, s =- 3, and s = 2 this yields A =- 2 / 3, B = 5 / 6, C =- 13 / 30, and D = 4 / 15. So, L- 1 { F ( s ) } ( t ) =- 2 3 L- 1 1 s ( t ) + 5 6 L- 1 1 s + 1 ( t ) 211 Chapter 7- 13 30 L- 1 1 s + 3 ( t ) + 4 15 L- 1 1 s- 2 ( t ) =- 2 3 + 5 6 e- t- 13 30 e- 3 t + 4 15 e 2 t ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part44 - Exercises 7.4...

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