5 20 points The linear system x 1 x 2 3 x 3 11 x 1 x 2 x 3 11 x 1 x 2 x 3 0 x 2

5 20 points the linear system x 1 x 2 3 x 3 11 x 1 x

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5. (20 points) The linear system x 1 + x 2 + 3 x 3 = 11 - x 1 + x 2 + x 3 = 11 x 1 - x 2 - x 3 = 0 x 2 + 2 x 3 = 11
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4 is inconsistent. Find the normal equations that determine a least squares solution to this system. (Do not solve them.) This linear system can be written as A~x = ~ b , where A = 1 1 3 - 1 1 1 1 - 1 - 1 0 1 2 and ~ b = 11 11 0 11 . The normal equations are A T A~x = A T ~ b . We have A T A = 1 - 1 1 0 1 1 - 1 1 3 1 - 1 2 1 1 3 - 1 1 1 1 - 1 - 1 0 1 2 = 3 - 1 1 - 1 4 7 1 7 15 and A T ~ b = 1 - 1 1 0 1 1 - 1 1 3 1 - 1 2 11 11 0 11 = 0 33 66 . Therefore the normal equations are 3 x 1 - x 2 + x 3 = 0 - x 1 + 4 x 2 + 7 x 3 = 33 x 1 + 7 x 2 + 15 x 3 = 66 . 6. (25 points) Find a general solution to the differential equation y 00 - 2 y 0 + y = 4 e t + 3 t . The characteristic polynomial is r 2 - 2 r + 1 = ( r - 1) 2 . A general solution to the associated homogeneous problem is c 1 e t + c 2 te t . Using the Method of Undetermined Coefficients to find a particular solution y p involves a trial solution of the form y p = At 2 e t + Bt + C . One has y 0 p = 2 Ate t + At 2 e t + B y 00 p = 2 Ae t + 4 Ate t + At 2 e t .
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5 Plugging this into the left-hand side of the differential equation gives y 00 p - 2 y 0 p + y p = 2 Ae t + 4 Ate t + At 2 e t - 4 Ate t - 2 At 2 e 5 - 2 B + At 2 e t + Bt + C = 2 Ae t + Bt + ( C - 2 B ) . Setting this equal to 4 e t + 3 t and equating coefficients gives 2 A = 4 , B = 3 , and C - 2 B = 0 ; therefore A = 2 , B = 3 , and C = 6 . This gives y p = 2 t 2 e t + 3 t + 6 , so a general solution to the nonhomogeneous problem is y = 2 t 2 e t + 3 t + 6 + c 1 e t + c 2 te t . 7. (16 points) Express the system x 00 + 3 x 0 + 2 x + 7 y = e t y 0 + x 0 + x - y = cos t x (0) = 3 , x 0 (0) = 5 , y (0) = - 1 as a matrix system in the form ~x 0 = A~x + ~ f , ~x (0) = ~x 0 . (Do not solve the system.) With the assignments in the left-hand column and the derivatives in the right: x 1 = x x 0 1 = x 0 = x 2 x 2 = x 0 x 0 2 = x 00 = e t - 3 x 0 - 2 x - 7 y = e t - 3 x 2 - 2 x 1 - 7 x 3 x 3 = y x 0 3 = y 0 = cos t - x 0 - x + y = cos t - x 2 - x 1 +
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