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141Bquiz9csols

141Bquiz9csols - 1 MATH 141B QUIZ#9 Solutions Date Friday 1...

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Unformatted text preview: 1 MATH 141B QUIZ#9 Solutions Date: Friday March 28, 2008 1. The auxiliary equation is r 2 + r- 12 = 0 , which factors as ( r- 3)( r + 4) = 0 . Therefore, r 1 =- 4 and r 2 = 3 , and hence y h = c 1 e − 4 x + c 2 e 3 x . Now, 1 = y (0) = c 1 + c 2 and since y ′ h =- 4 c 1 e − 4 x- 3 c 2 e 3 x , we get- 1 = y ′ h (0) =- 4 c 1 + 3 c 2 . Solving the linear system braceleftBigg c 1 + c 2 = 1- 4 c 1 + 3 c 2 =- 1 gives c 1 = 4 / 7 and c 2 = 3 / 7 . Therefore, D) is the answer. 2. The auxiliary equation is ( r- 3) 2 = 0 , so r = 3 is a repeated root and y h = c 1 e 3 x + c 2 xe 3 x . Since r = 0 is not a root, and the right hand side of the differential equation has degree zero, y p has the form y p = A , which is an arbitrary polynomial of degree zero. Since y ′ p = 0 and y ′′ p = 0, putting these into the differential equation gives 9 A = 3 so A = 1 / 3 . Therefore, y = y p + y h = 1 3 + c 1 e 3 x + c 2 xe 3 x is the general solution to the given non-homogeneous differential equation. Now apply the initial conditions.differential equation....
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141Bquiz9csols - 1 MATH 141B QUIZ#9 Solutions Date Friday 1...

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