# sol7 - x 1 Thus a general solution to our equation is also...

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MAP 2302 Fall-2011 Quiz 7 1. Solve the initial value problem y 000 - y 00 - 4 y 0 + 4 y = 0; y (0) = - 4, y 0 (0) = - 1, y 00 (0) = - 19. SOLUTION. The characteristic equation is r 3 - r 2 - 4 r +4 = 0 has a root r = 1. Then r 3 - r 2 - 4 r + 4 = ( r - 1)( r 2 + ar + b ). Comparing coeﬃcients on the left and the right we can ﬁnd that a = 0 b = - 4. Thus, the other two roots are r = 2 and r = - 2. The general solution is y = C 1 e t + C 2 e 2 t + C 3 e - 2 t . Then y 0 = C 1 e t + 2 C 2 e 2 t - 2 C 3 e - 2 t and y 00 = C 1 e t + 4 C 2 e 2 t + 4 C 3 e - 2 t . The initial conditions gives us the system C 1 + C 2 + C 3 = - 4 C 1 + 2 C 2 - 2 C 3 = - 1 C 1 + 4 C 2 + 4 C 3 = - 19. We subtract the ﬁrst equation from the rest to obtain the system C 2 - 3 C 3 = 3 3 C 2 + 3 C 3 = - 15 which can be solved for C 2 by adding these two up. Thus, C 2 = - 3. Hence C 3 = - 2 and C 1 = 1. The answer is y = e t - 3 e 2 t - 2 e - 2 t . 2. Use the annihilator method to determine the form of a particular solution for the equation u 00 - 5 u 0 + 6 u = cos 2 x + 1 . 1

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SOLUTION: The equation in the operator form is ( D 2 - 5 D + 6)[ u ] = cos 2 x + 1 . The function cos 2 x is annihilated by ( D 2 +4). The constant 1 is annihilated by by D . Hence D ( D 2 + 4) annihilates cos 2
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Unformatted text preview: x + 1. Thus a general solution to our equation is also a solution to the homogeneous equation D ( D 2 + 4)( D 2-5 D + 6)[ u ] = 0 . The characteristic equation r ( r 2 +4)( r 2-5 r +6) = 0 has the following roots: r = 2 non-repeated the fundamental solution e 2 x ; r = 3 non-repeated the fundamental solution e 3 x ; r = 0 non-repeated the fundamental solution 1; r = ± 2 i complex conjugate the fund. sol. are cos 2 x and sin 2 x . So, the fundamental solution set is e 2 x , e 3 x , 1, cos 2 x , and sin 2 x . A general solution is u = C 1 e 2 x + C 2 e 3 x + C 3 + C 4 cos 2 x + C 5 sin 2 x = u h + u p . Since the solution to homogeneous equation u 00-5 u + 6 u = 0 is u h = C 1 e 2 x + C 2 e 3 x , we obtain u p = C 3 + C 4 cos 2 x + C 5 sin 2 x. 2...
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## This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.

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sol7 - x 1 Thus a general solution to our equation is also...

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