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# 527ex1sl - 642:527 EXAM 1 SOLUTIONS 3n(x 2)2n converge n2...

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642:527 EXAM 1 SOLUTIONS FALL 2009 1. For what values of x does the power series s n =0 3 n ( x 2) 2 n n 2 converge? Solution: We use the ratio test: lim n →∞ v v v v a n +1 a n v v v v = lim n →∞ v v v v 3 n +1 ( x + 2) 2 n +2 / ( n + 1) 2 3 n ( x + 2) 2 n /n 2 v v v v = 3 | x + 2 | 2 lim n →∞ p n n + 1 P 2 = 3 | x + 2 | 2 . The series converges if this limit is less than 1, i.e., if | x + 2 | < 1 / 3, that is, for all x with 2 1 / 3 < x < 2 + 1 / 3. 2. (a) What will be the form of the partial fraction expansion of G ( s ) = s 2 + 7 ( s 2)( s 2 + 9) ? Your answer will involve some constants which you should not evaluate . Solution: G ( s ) = A s 2 + Bs + C s 2 + 9 . (b) Find the inverse Laplace transform of e 5 s [ G ( s ) + 1]. Your answer may be left in terms of the constants you introduced in (a). Solution: From the table, G ( s ) + 1 = L{ g ( t ) } , where g ( t ) = Ae 2 t + B cos 3 t + ( C/ 3)sin 3 t + δ ( t ). Then L 1 { e 5 s [ G ( s ) + 1] } = H ( t 5) g ( t 5) = H ( t 5) b Ae 2( t 5) + B cos 3( t 5) + ( C/ 3)sin 3( t 5) + δ ( t 5) B . 3. A function f ( t ) is de±ned for t 0 by f ( t ) = ± e 2 t , if 2 t < 5, 0 , if 0 t < 2 or t 5. (a) Express f ( t ) in terms of a single formula using the Heaviside function, then ±nd its Laplace transform. Solution: We have f ( t ) = H ( t 2) e 2 t H ( t 5) e 2 t = H ( t 2) e 4 e 2( t 2) H ( t 5) e 10 e 2( t 5) . and since L{ e 2 t } = 1 / ( s + 2) the Laplace transform is F ( s ) = e 4 e 2 s s + 2 e 10 e 5 s s + 2 (b) Solve the initial value problem: x ( t ) 2 x ( t ) = f ( t ), x (0) = 3. Solution: The Laplace transform X ( s ) if the solution satis±es sX ( s ) 3 2 X (3) = F ( s ) or X ( s ) = 3 s 2 + e 4 e 2 s s 2 4 e 10 e 5 s s 2 4 Since L 1 [2 / ( s 2 4)] = sinh(2 t ) we have x ( t ) = 3 e 2 t + e 4 2 H ( t 2)sinh 2( t 2) e 10 2 H ( t 5)sinh 2( t 5) .

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527ex1sl - 642:527 EXAM 1 SOLUTIONS 3n(x 2)2n converge n2...

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