This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 nonhomogeneous differential equation. Now apply the initial conditions.differential equation....
View Full
Document
 Spring '08
 MARCFABBRI
 Logic, Factors

Click to edit the document details