nagle_differential_equations_ISM_Part46

# nagle_differential_equations_ISM_Part46 - Exercises 7.5 32....

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Exercises 7.5 32. Applying the Laplace transform to both sides the given equation yields ± s 2 Y ( s ) - as - b ² - 5 [ sY ( s ) - a ] + 6 Y ( s ) = L ³ - 6 te 2 t ´ ( s ) = - 6 ( s - 2) 2 ( s 2 - 5 s + 6 ) Y ( s ) = as + b - 5 a - 6 ( s - 2) 2 = as 3 + ( b - 9 a ) s 2 + (24 a - 4 b ) s + (4 b - 20 a - 6) ( s - 2) 2 Y ( s ) = as 3 + ( b - 9 a ) s 2 + (24 a - 4 b ) s + (4 b - 20 a - 6) ( s - 2) 2 ( s 2 - 5 s + 6) = as 3 + ( b - 9 a ) s 2 + (24 a - 4 b ) s + (4 b - 20 a - 6) ( s - 2) 3 ( s - 3) = A ( s - 2) 3 + B ( s - 2) 2 + C s - 2 + D s - 3 . (For the Laplace transforms of y 0 and y 00 we have used equations (7.8).) Solving for A , B , C , and D , we ﬁnd that A = 6 , B = 6 , C = 3 a - b + 6 , D = b - 2 a - 6 . Hence, Y ( s ) = 6 ( s - 2) 3 + 6 ( s - 2) 2 + 3 a - b + 6 s - 2 + b - 2 a - 6 s - 3 y ( t ) = 3 t 2 e 2 t + 6 te 2 t + (3 a - b + 6) e 2 t + ( b - 2 a - 6) e 3 t . 34. By Theorem 6 in Section 7.3, L ³ t 2 y 00 ( t ) ´ ( s ) = ( - 1) 2 d 2 ds 2 [ L{ y 00 ( t ) } ( s )] = d 2 ds 2 [ L{ y 00 ( t ) } ( s )] . (7.9) Theorem 5 in Section 7.3 says that L{ y 00 ( t ) } ( s ) = s 2 Y ( s ) - y (0) s - y 0 (0) . Substituting this equation into (7.9) yields L ³ t 2 y 0 ( t ) ´ ( s ) = d 2 ds 2 ± s 2 Y ( s ) - y (0) s - y 0 (0) ² = d 2 ds 2 ± s 2 Y ( s ) ² = d ds ± s 2 Y 0 ( s ) + 2 sY ( s ) ² = s 2 Y 00 ( s ) + 4 sY 0 ( s ) + 2 Y ( s ) . 36. We apply the Laplace transform to the given equation and obtain L{ ty 00 } ( s ) - L{ ty 0 } ( s ) + L{ y } ( s ) = L{ 2 } ( s ) = 2 s . (7.10) 221

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Chapter 7 Using Theorem 5 in Section 7.3 and the initial conditions, we express L{ y 00 } and L{ y 0 } in terms of Y . L{
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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_Part46 - Exercises 7.5 32....

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