276_pdfsam_math 54 differential equation solutions odd

276_pdfsam_math 54 differential equation solutions odd - 1...

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Chapter 5 Substitution of this solution into (5.12) and (5.13) yield z = 1 4 ( x + 4 x ) = c 1 e 8 t + c 3 , y = 1 16 ( x 8 x + 8 x ) = 1 2 ( c 1 e 8 t c 2 e 4 t + c 3 ) . 29. We begin by expressing the system in operator notation ( D λ )[ x ] + y = 0 , 3 x + ( D 1)[ y ] = 0 . We eliminate y by applying ( D 1) to the first equation and subtracting the second equation from it. This gives { ( D 1)( D λ ) ( 3) } [ x ] = 0 D 2 ( λ + 1) D + ( λ + 3) [ x ] = 0 . (5.14) Note that since the given system is homogeneous, y ( t ) also satisfies this equation (compare (7) and (8) on page 247 of the text). So, we can investigate solutions x ( t ) only. The auxiliary equation, r 2 ( λ + 1) r + ( λ + 3) = 0, has roots r 1 = ( λ + 1)
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Unformatted text preview: + 1) + √ ∆ 2 , where the discriminant ∆ := ( λ + 1) 2 − 4( λ + 3). We consider two cases: i) If λ + 3 < 0, i.e. λ < − 3, then ∆ > ( λ + 1) 2 and the root r 2 > ( λ + 1) + | λ + 1 | 2 = 0 . Therefore, the solution x ( t ) = e r 2 t is unbounded as t → + ∞ . ii) If λ + 3 ≥ 0, i.e. λ ≥ − 3, then ∆ ≤ ( λ + 1) 2 . If ∆ < 0, then a fundamental solution set to (5.14) is e ( λ +1) t/ 2 cos ³ √ − ∆ t 2 ´ , e ( λ +1) t/ 2 sin ³ √ − ∆ t 2 ´ . (5.15) 272...
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