hw7solns - Section 3.4 5c(D2 4D 5(x3 cos 2x = D2(x3 cos 2x...

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Section 3.4 5c. ( D 2 - 4 D + 5)( x 3 - cos 2 x ) = D 2 ( x 3 - cos 2 x ) - 4 D ( x 3 - cos 2 x ) + 5( x 3 - cos 2 x ) = (6 x + 4 cos 2 x ) - 4(3 x 2 + 2 sin 2 x ) + (5 x 3 - 5 cos 2 x ) 7. [ D 3 + 2 D ]( D 2 - D )( x 6 - 2 x 2 ) = [ D 3 + 2 D ]( D 6 x 5 - Dx 4 - 6 x 5 + 4 x ) = ( D 3 + 2 D )(30 x 4 - 4 - 6 x 5 + 4 x ) = 720 x - 0 - 360 x 2 - 0 + 240 x 3 - 0 - 60 x 4 + 8 = - 60 x 4 + 240 x 3 - 360 x 2 + 720 x + 8 ( D 2 - D )( D 3 + 2 D )( x 6 - 2 x 2 ) = ( D 2 - D )(120 x 3 - 0 + 12 x 5 - 8 x ) = (720 x + 240 x 3 - 360 x 2 - 60 x 4 + 8) Both P ( D ) Q ( D )( y ) and Q ( D ) P ( D )( y ) yield the same answer. 15. D 3 ( D - 2)( D + 3)( x 2 - 2 x + 7) = ( D - 2)( D + 3) D 3 ( x 2 - 2 x + 7) = 0 21b. y + 3 y - 4 y = 0 ( D 2 + 3 D - 4)( y ) = 0 ( D + 4)( D - 1)( y ) = 0 Section 3.5 2. We have y + 4 y - 3 y = 0, y (0) = 0, y (0) = 1. (i) y = - 4 7 y + 3 7 y . Let u 1 = y , u 2 = y . Then u 1 = u 2 u 2 = - 4 7 u 2 + 3 7 u 1 and u 1 (0) = 0 u 2 (0) = 1 (ii) step size 1 .2 .01 solution at x = 5 5.9589 5.9619 5.9619 4. We have y + x 2 y + 12 y = 0, y (0) = 0, y (0) = 7. Solve for the highest derivative: y = - x 2 y - 12 y . Let u 1 = y , u 2 = y . Then u 1 = u 2 u 2 = - x 2 u 2 - 12 u 1 with u 1 (0) = 0 , u 2 (0) = 7 Then y (5) = . 0104 with h = . 01. 5. We have y + 4 y + 3 sin y = 1, y (0) = - 1, y (0) = π . Solve for y : y = - 4 y - 3 sin y + 1. Let u 1 = y , u 2 = y . Then u 1 = u 2 u 2 = - 4 u 2 - 3 sin u 1 + 1 with u 1 (0) = - 1 , u 2 (0) = π Then y (5) = . 6846 with h = . 01.
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11. We have 8 y + y = 0,
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  • Winter '07
  • brown
  • Characteristic polynomial, Berlin U-Bahn, general solution, linearly independent solutions, R2-D2

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