286_pdfsam_math 54 differential equation solutions odd

286_pdfsam_math 54 - Chapter 5(b r =(c r = 1 9 12 4(6 1 5 = = 3 2 D 2 D 6 =(D 3(D 2 2 2 92 4(5)2 9 11 = = 5 1/2 2D 2 9D 5 =(D 5(2D 1 4 4 2 D 2 =(D

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Chapter 5 (b) r = 1 ± p 1 2 4( 6) 2 = 1 ± 5 2 = 3 , 2 D 2 + D 6=( D +3)( D 2). (c) r = 9 ± p 9 2 4( 5)2 4 = 9 ± 11 4 = 5 , 1 / 2 2 D 2 +9 D 5=( D + 5)(2 D 1). (d) r = ± 2 D 2 2=( D + 2)( D 2). EXERCISES 5.3: Solving Systems and Higher–Order Equations Numerically, page 261 1. We isolate y 0 ( t ) frst and obtain an equivalent equation y 0 ( t )=3 y ( t ) ty 0 ( t )+ t 2 . Denoting x 1 := y , x 2 := y 0 we conclude that x 0 1 = y 0 = x 2 , x 0 2 =( y 0 ) 0 = y 0 =3 y ty 0 + t 2 x 1 tx 2 + t 2 , with initial conditions x 1 (0) = y (0) = 3, x 2 (0) = y 0 (0) = 6. ThereFore, given initial value problem is equivalent to x 0 1 = x 2 , x 0 2 x 1 tx 2 + t 2 , x 1 (0) = 3 ,x 2 (0) = 6 . 3. Isolating y (4) ( t ), we get y (4) ( t )=
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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