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Chapter 7

Chapter 7 - SECTION 7.2 1 t By definition we have L t = Z...

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Unformatted text preview: SECTION 7.2 1. t By definition, we have L { t } = Z ∞ e- st t dt L { t } = lim u-→∞ Z u e- st t dt. Using integration by parts L { t } = lim u-→∞ [- ( t/s ) e- st | u + (1 /s ) lim u-→∞ Z u e- st dt ] L { t } = lim u-→∞ [- ( t/s ) e- st | u- (1 /s 2 ) e- st | u ] L { t } = lim u-→∞ [- ( u/s ) e- su- (1 /s 2 ) e- su + (1 /s 2 )] . We know lim u-→∞ e- su = 0 , s > 0 and using L’Hospital’s Rule lim u-→∞ ue- su = lim u-→∞ u e su = lim u-→∞ 1 se su = 0 , s > . Hence we have L { t } = (1 /s 2 ) , s > . 3. e 6 t By definition, we have L { e 6 t } = Z ∞ e- st e 6 t dt = Z ∞ e (6- s ) t dt L { e 6 t } = lim u-→∞ Z u e (6- s ) t dt L { e 6 t } = lim u-→∞ 1 6- s e (6- s ) t | u 1 L { e 6 t } = lim u-→∞ 1 6- s ( e (6- s ) u- 1) L { e 6 t } = 1 s- 6 , s > 6 . 5. cos 2 t By definition, we have L { cos 2 t } = Z ∞ e- st cos 2 t dt L { cos 2 t } = lim u-→∞ Z u e- st cos 2 t dt. Using integration by parts Z e- st cos 2 t dt = (1 / 2) e- st sin 2 t + Z ( s/ 2) e- st sin 2 t dt. Again applying integration by parts Z e- st cos 2 t dt = (1 / 2) e- st sin 2 t + ( s/ 2)[- (1 / 2) e- st cos 2 t- Z ( s/ 2) e- st cos 2 t dt ] Z e- st cos 2 t dt = (1 / 2) e- st sin 2 t- ( s/ 4) e- st cos 2 t- ( s 2 / 4) Z e- st cos 2 t dt ] (1 + s 2 / 4) Z e- st cos 2 t dt = (1 / 2) e- st sin 2 t- ( s/ 4) e- st cos 2 t 4 + s 2 4 Z e- st cos 2 t dt = (1 / 2) e- st sin 2 t- ( s/ 4) e- st cos 2 t Z e- st cos 2 t dt = 4 4 + s 2 [(1 / 2) e- st sin 2 t- ( s/ 4) e- st cos 2 t ] so that L { cos 2 t } = lim u-→∞ 4 4 + s 2 [(1 / 2) e- st sin 2 t- ( s/ 4) e- st cos 2 t ] | u L { cos 2 t } = lim u-→∞ 4 4 + s 2 [((1 / 2) e- su sin 2 u- ( s/ 4) e- su cos 2 u )- (- ( s/ 4))] L { cos 2 t } = 4 4 + s 2 s 4 = s s 2 + 4 , s > . 2 7. e 2 t cos 3 t By definition, we have L { e 2 t cos 3 t } = Z ∞ e- st e 2 t cos 3 t dt L { e 2 t cos 3 t } = lim u-→∞ Z u e (2- s ) t cos 3 t dt. Using integration by parts Z e (2- s ) t cos 3 t dt = 1 3 e (2- s ) t sin 3 t- Z 2- s 3 e (2- s ) t sin 3 t dt. Again applying integration by parts Z e (2- s ) t cos 3 t dt = 1 3 e (2- s ) t sin 3 t- 2- s 3 [- 1 3 e (2- s ) t cos 3 t + Z 2- s 3 e (2- s ) t cos 3 t dt ] Z e (2- s ) t cos 3 t dt = 1 3 e (2- s ) t sin 3 t + 2- s 9 e (2- s ) t cos 3 t- ( 2- s 3 ) 2 Z e (2- s ) t cos 3 t dt (1 + ( 2- s 3 ) 2 ) Z e (2- s ) t cos 3 t dt = 1 3 e (2- s ) t sin 3 t + 2- s 9 e (2- s ) t cos 3 t 9 + ( s- 2) 2 9 Z e (2- s ) t cos 3 t dt = 1 3 e (2- s ) t sin 3 t + 2- s 9 e (2- s ) t cos 3 t Z e (2- s ) t cos 3 t dt = 3 9 + ( s- 2) 2 e (2- s ) t sin 3 t + 2- s 9 + ( s- 2) 2 e (2- s ) t cos 3 t so that L { e 2 t cos 3 t } = lim u-→∞ 3 9 + ( s- 2) 2 e (2- s ) t sin 3 t + 2- s 9 + ( s- 2) 2 e (2- s ) t cos 3 t | u L { e 2 t cos 3 t } = lim u-→∞ ( 3 9 + ( s- 2) 2 e (2- s ) u sin 3 u + 2- s 9 + ( s- 2) 2 e (2- s ) u cos 3 u )- 2- s 9 + ( s- 2) 2 L { e 2 t cos 3 t } = s- 2 9 + ( s- 2) 2 , s > 2 ....
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Chapter 7 - SECTION 7.2 1 t By definition we have L t = Z...

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