quiz3 - Quiz 3 Math 240 - Calculus III February 17, 2009...

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Quiz 3 Name: Math 240 - Calculus III February 17, 2009 Note: In order to receive full credit, you must show work that justifies your answer. 1. Find the general solution of the following second-order differential equation: y 00 - 9 y = 0 Solution : The auxiliary equation is m 2 - 9 m = 0, which has solutions m = - 3 , 3. This implies that y 1 ( x ) = e - 3 x and y 2 ( x ) = e 3 x are both solutions to the given differential equation. Therefore, the general solution is y ( x ) = c 1 e - 3 x + c 2 e 3 x . 2. Find the general solution of the given system: dx dt = 2 x + y, and dy dt = 4 y. Solution : This system corresponds to the matrix equation X 0 = ± 2 1 0 4 ² X . Since the matrix is diagonal, its eigenvalues are λ 1 = 2 and λ 2 = 4. An eigenvector for λ 1 = 2 is v 1 = h 1 , 0 i , and an eigenvector for λ 2 = 4 is v 2 = h 1 , 2 i . Therefore, the general solution is X = c 1 ± 1 0 ² e 2 t + c 2 ± 1 2 ² e 4 t .
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Quiz 3 Name: Math 240 - Calculus III February 19, 2009 Note: In order to receive full credit, you must show work that justifies your answer.
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quiz3 - Quiz 3 Math 240 - Calculus III February 17, 2009...

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