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286_pdfsam_math 54 differential equation solutions odd

286_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 (b) r = 1 ± 1 2 4( 6) 2 = 1 ± 5 2 = 3 , 2 D 2 + D 6 = ( D + 3)( D 2). (c) r = 9 ± 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 ( t ) first and obtain an equivalent equation y ( t ) = 3 y ( t ) ty ( t ) + t 2 . Denoting x 1 := y , x 2 := y we conclude that x 1 = y = x 2 , x 2 = ( y ) = y = 3 y ty + t 2 = 3 x 1 tx 2 + t 2 , with initial conditions x 1 (0) = y (0) = 3, x 2 (0) = y (0) = 6. Therefore, given initial value problem is equivalent to x 1 = x 2 , x 2 = 3 x 1 tx 2 + t 2 , x 1 (0) = 3 , x 2 (0) = 6 . 3. Isolating
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