nagle_differential_equations_ISM_Part41

nagle_differential_equations_ISM_Part41 - Chapter 6 8 c 1 x...

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Unformatted text preview: Chapter 6 8. c 1 x + c 2 x ln x + c 3 x 3- x 2 10. 1 / (10 x ) R g ( x ) dx + ( x 4 / 15) R x- 5 g ( x ) dx- ( x/ 6) R x- 2 g ( x ) dx REVIEW PROBLEMS 2. (a) Linearly independent. (b) Linearly independent. (c) Linearly dependent. 4. (a) c 1 e- 3 x + c 2 e- x + c 3 e x + c 4 xe x (b) c 1 e x + c 2 e (- 2+ √ 5 ) x + c 3 e (- 2- √ 5 ) x (c) c 1 e x + c 2 cos x + c 3 sin x + c 4 x cos x + c 5 x sin x (d) c 1 e x + c 2 e- x + c 3 e 2 x- ( x/ 2) e x + ( x/ 2) + (1 / 4) 6. c 1 e- x/ √ 2 cos ( x/ √ 2 ) + c 2 e- x/ √ 2 sin ( x/ √ 2 ) + c 3 e x/ √ 2 cos ( x/ √ 2 ) + c 4 e x/ √ 2 sin ( x/ √ 2 ) + sin ( x 2 ) 8. (a) c 1 xe- x + c 2 + c 3 x + c 4 x 2 (b) c 1 xe- x + c 2 x 2 e- x (c) c 1 + c 2 x + c 3 x 2 + c 4 cos 3 x + c 5 sin 3 x (d) c 1 cos x + c 2 sin x + c 3 x cos x + c 4 x sin x 10. (a) c 1 x 1 / 2 + c 2 x- 1 / 2 + c 3 x (b) c 1 x- 1 + c 2 x cos ( √ 3 ln x ) + c 3 x sin ( √ 3 ln x ) 196 CHAPTER 7: Laplace Transforms EXERCISES 7.2: Definition of the Laplace Transform 2. For s > 0, using Definition and integration by parts twice, we compute L t 2 ( s ) = ∞ Z e- st t 2 dt = lim N →∞ N Z e- st t 2 dt = lim N →∞ - t 2 e- st s N + 2 s N Z te- st dt = lim N →∞- t 2 s- 2 t s 2- 2 s 3 e- st N = lim N →∞ 2 s 3- N 2 s + 2 N s 2 + 2 s 3 e- sN = 2 s 3 , because, for s > 0 and any k , N k e- sN → 0 as N → ∞ ....
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nagle_differential_equations_ISM_Part41 - Chapter 6 8 c 1 x...

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