chapter6.page03

# chapter6.page03 - -s-1 = b e t ( s ) = 1 s-1 Solve this...

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1. THE LAPLACE TRANSFORM 3 f = L - 1 ( b f ) b f f = L - 1 ( b f ) b f 1 1 s ( s > 0) t 1 s 2 ( s > 0) t n ( n 0) n ! s n +1 ( s > 0) t a ( a > - 1) Γ( a +1) s a +1 ( s > 0) e at 1 s - a ( s > 0) sin( kt ) k s 2 + k 2 ( s > 0) cos( kt ) s s 2 + k 2 ( s > 0) sinh( kt ) k s 2 - k 2 ( s > | k | ) cosh( kt ) s s 2 - k 2 ( s > | k | ) u ( t - a ) e as s ( s > 0) Example 1.1 . Find b x ( s ) if x 00 +2 x 0 - 3 x = e t ,x (0) = 1 , x 0 (0) = - 1 . Solution Apply Laplace transform on both side of the equation, L ( x 00 + 2 x 0 - 3 x )( s ) = L ( e t )( s ) Using the linear property \ af + bg ( s ) = a b f ( s ) + b b g ( s ) we have L ( x 00 + 2 x 0 - 3 x )( s ) = b x 00 ( s ) + 2 b x 0 ( s ) - 3 b x ( s ) . Together with b x 00 ( s ) = s 2 b x ( s ) - sx (0) - x 0 (0) and b x 0 ( s ) = s b x ( s ) - x (0) We have, due to x (0) = 1 , x 0 (0) = - 1 , L ( x 00 + 2 x 0 - 3 x )( s ) = b x 00 ( s ) + 2 b x 0 ( s ) - 3 b x ( s ) = [ s 2 b x ( s ) - sx (0) - x 0 (0)] + 2[ s b x ( s ) - x (0)] - 3 b x ( s ) = ( s 2 + 2 s - 3) b x ( s ) - s - 1 So we have an algebraic equation for b x ( s ) , ( s 2 + 2 s - 3) b x ( s )
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Unformatted text preview: -s-1 = b e t ( s ) = 1 s-1 Solve this equation for b x ( s ) , ( s 2 + 2 s-3) b x ( s )-s-1 = b e t ( s ) = 1 s-1 ( s 2 + 2 s-3) b x ( s ) = 1 s-1 + s + 1 ( s 2 + 2 s-3) b x ( s ) = 1 s-1 + ( s +1)( s-1) s-1 ( s 2 + 2 s-3) b x ( s ) = 1+( s +1)( s-1) s-1 ( s 2 + 2 s-3) b x ( s ) = 1+( s 2-1) s-1 ( s 2 + 2 s-3) b x ( s ) = s 2 s-1...
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## This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.

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