399_pdfsam_math 54 differential equation solutions odd

# 399_pdfsam_math 54 differential equation solutions odd -...

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Exercises 7.2 (b) Note that s ( a + ib )=( s a ) ib . Multiplying the result in (a) by the complex conjugate of the denominator, that is, ( s a )+ bi ,weget 1 s ( a + ib ) = ( s a )+ ib [( s a ) ib ] · [( s a )+ ib ] = ( s a )+ ib ( s a ) 2 + b 2 , where we used the fact that, for any complex number z , z z = | z | 2 . (c) From (a) and (b) we klnow that L ± e ( a + ib ) t ² ( s )= ( s a )+ ib ( s a ) 2 + b 2 . Writing ( s a )+ ib ( s a ) 2 + b 2 = s a ( s a ) 2 + b 2 + b ( s a ) 2 + b 2 i, we see that Re ³ L ± e ( a + ib ) t ² ( s ) ´ =Re µ s a ( s a ) 2 + b 2 + b ( s a ) 2 + b 2 i = s a ( s a ) 2 + b 2 , (7.2) Im ³ L ± e ( a + ib ) t ² ( s ) ´ =Im µ s a ( s a ) 2 + b 2 + b ( s a ) 2 + b 2 i = b ( s a ) 2 + b 2 . (7.3) On the other hand, by Euler’s formulas, Re ³ e st e ( a + ib ) t ´ = e st Re ³ e at (cos bt + i sin bt ) ´ = e st e at cos bt and so Re ³ L ± e ( a + ib ) t ² ( s ) ´ =R e
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Unformatted text preview: ∞ Z e − st e ( a + ib ) t dt = Re ∞ Z e − s e ( a + ib ) t dt = ∞ Z Re ³ e − s e ( a + ib ) t ´ dt = ∞ Z e − st e at cos bt dt = L ± e at cos bt ² ( s ) , which together with (7.2) gives the last entry in Table 7.1. Similarly, Im ³ L ± e ( a + ib ) t ² ( s ) ´ = L ± e at sin bt ² ( s ) , and so (7.3) gives the Laplace transform of e at sin bt . 395...
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