QUIZ4solns - MATH 527 QUIZ 4 Name: Section: Date: Nov. 29,...

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Unformatted text preview: MATH 527 QUIZ 4 Name: Section: Date: Nov. 29, 2007 BE SURE TO SHOW ALL OF YOUR WORK. 1. (10 pts) Solve the following initial value problem using the Laplace Trans- form. y + 5 y + 6 y = u 3 ( t ) y (0) = 0 y (0) = 0 Solution: L{ y + 5 y + 6 y } = L{ u 3 ( t ) } By the linearity of the Laplace Transform L{ y } + 5 L{ y } + 6 L{ y } = L{ u 3 ( t ) } By theorem 6.2.1 from Boyce & Diprima s 2 L{ y }- sy (0)- y (0) + 5( s L{ y }- y (0)) + 6 L{ y } = e 3 s s Plug in our initial conditions s 2 L{ y } + 5 s L{ y } + 6 L{ y } = e 3 s s Solve for L{ y } L{ y } ( s 2 + 5 s + 6) = e 3 s s Y ( s ) = e 3 s s ( s + 3)( s + 2) Now, by partial fractions 1 s ( s + 3)( s + 2) = A s + B ( s + 3) + C ( s + 2) 1 = A ( s + 3)( s + 2) + Bs ( s + 2) + Cs ( s + 3) Let s =- 2 , then 1 = C (- 2)(1) = C =- 1 / 2 , then for s =- 3 , we get 1 = B (- 3)(- 1) = B = 1 / 3 , and finally s = 0 , gives 1 = A (3)(2) = A = 1 / 6 ....
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This homework help was uploaded on 04/08/2008 for the course MATH 527 taught by Professor Boucher during the Fall '07 term at New Hampshire.

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QUIZ4solns - MATH 527 QUIZ 4 Name: Section: Date: Nov. 29,...

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