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nagle_differential_equations_ISM_Part44

# nagle_differential_equations_ISM_Part44 - Exercises 7.4...

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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

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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 . 30. Solving for F ( s ) yields F ( s ) = 2 s + 5 ( s - 1)( s 2 + 2 s + 1) = A ( s + 1) 2 + B s + 1 + C s - 1 . (7.6) Thus, clearing fractions, we conclude that 2 s + 5 = A ( s - 1) + B ( s 2 - 1 ) + C ( s + 1) 2 .
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nagle_differential_equations_ISM_Part44 - Exercises 7.4...

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