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Chem Differential Eq HW Solutions Fall 2011 20

Chem Differential Eq HW Solutions Fall 2011 20 - p = π 25...

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20 Chapter 2 Fourier Series fact that e z + w = e z e w ), we obtain I 1 + iI 2 = 1 a + ib e x ( a + ib ) + C = ( a + ib ) ( a + ib ) · ( a + ib ) e ax e ibx + C = a - ib a 2 + b 2 e ax ( cos bx + i sin bx ) + C = e ax a 2 + b 2 ( a cos bx + b sin bx ) + i ( - b cos bx + a sin bx ) + C. (c) Equating real and imaginary parts in (b), we obtain I 1 = e ax a 2 + b 2 ( a cos bx + b sin bx ) and I 2 = e ax a 2 + b 2 ( - b cos bx + a sin bx ) . 21. By Exercise 19, 2 π 0 ( e it + 2 e - 2 it ) dt = 1 i e it + 2 - 2 i e - 2 it 2 π 0 = - i =1 e 2 πi + i =1 e - 2 πi - ( - i + i ) = 0 . Of course, this result follows from the orthogonality relations of the complex expo- nential system (formula (11), with
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Unformatted text preview: p = π ). 25. First note that 1 + it 1-it = (1 + it ) 2 (1-it )(1 + it ) = 1-t 2 + 2 it 1 + t 2 = 1-t 2 1 + t 2 + i 2 t 1 + t 2 . Hence ³ 1 + it 1-it dt = ³ 1-t 2 1 + t 2 dt + i ³ 2 t 1 + t 2 dt = ³ (-1 + 2 1 + t 2 dt + i ³ 2 t 1 + t 2 dt =-t + 2 tan-1 t + i ln(1 + t 2 ) + C....
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