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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.
 Spring '08
 MAZMANI
 Equations

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