quiz4 - Quiz 4 Math 240 - Calculus III February 24, 2009...

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Quiz 4 Name: Math 240 - Calculus III February 24, 2009 Note: In order to receive full credit, you must show work that justifies your answer. 1. (6 points) Find the general solution of the following system of differential equations: X 0 = ± 1 3 3 1 ² X + ± 2 - 3 ² e 3 t Solution : The matrix ± 1 3 3 1 ² has eigenvalues λ 1 = - 2 and λ 2 = 4, with corresponding eigenvectors v 1 = h- 1 , 1 i and v 2 = h 1 , 1 i , respectively. The complementary solution is then X c = c 1 ± - 1 1 ² e - 2 t + c 2 ± 1 1 ² e 4 t . We suppose a particular solution has the form X p = ± a b ² e 3 t . Substituting X p into the differential equation, we find that a = 1 and b = 0. The general solution is thus X = c 1 ± - 1 1 ² e - 2 t + c 2 ± 1 1 ² e 4 t + ± 1 0 ² e 3 t . 2. (2 points each) Find differential operators that “annihilate” the following functions: (a) y ( x ) = e 5 x Solution : The annihilator is D - 5 since ( D - 5) y = y 0 - 5 y = 5 e 5 x - 5 e 5 x = 0 . (b)
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quiz4 - Quiz 4 Math 240 - Calculus III February 24, 2009...

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