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QUIZ4solns

# QUIZ4solns - MATH 527 Name BE QUIZ 4 Section Date Nov 29...

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MATH 527 QUIZ 4 Name: Section: Date: Nov. 29, 2007 B E 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 . Y ( s ) = e 3 s parenleftbigg 1 s + 1 s + 3 + 1 s + 2 parenrightbigg . L 1 { Y ( s ) } = y ( t ) = L 1 braceleftbigg e 3 s parenleftbigg 1 6 1 s + 1 3 1 s + 3 + - 1 2 1 s + 2 parenrightbiggbracerightbigg . y ( t ) = u 3 ( t ) parenleftbigg 1 6 + 1 3 e 3( t 3) + - 1 2 e 2( t 3) parenrightbigg .

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2. (10 pts) Convert the following linear system into matrix form and solve using gaussian elimination. 2 x + 4 y + 3 z = 1 - x + y + 2 z = - 1 4 x + 2 y - 2 z = 0 Solution: As a matrix equation we have
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QUIZ4solns - MATH 527 Name BE QUIZ 4 Section Date Nov 29...

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