395_pdfsam_math 54 differential equation solutions odd

395_pdfsam_math 54 differential equation solutions odd -...

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Exercises 7.2 L ± t 2 ² ( s )= 2! s 2+1 = 2 s 3 , L{ t } ( s 1! s 1+1 = 1 s 2 , 1 } ( s 1 s ,s > 0 . Thus the formula L ± 6 e 3 t t 2 +2 t 8 ² ( s )=6 1 s +3 2 s 3 1 s 2 8 1 s = 6 s 2 s 3 + 2 s 2 8 s , is valid for s in the intersection of the sets s> 3and 0, which is 0. 15. Using the linearity of Laplace transform and Table 7.1 on page 358 in the text, we get L ± t 3 te t + e 4 t cos t ² ( s L ± t 3 ² ( s ) −L ± te t ² ( s )+ L ± e 4 t cos t ² ( s ) = 3! s 3+1 1! ( s 1) 1+1 + s 4 ( s 4) 2 +1 2 = 6 s 4 1 ( s 1) 2 + s 4 ( s 4) 2 , which is valid for 4. 17. Using the linearity of Laplace transform and Table 7.1 on page 358 in the text, we get L ± e 3 t sin 6 t t 3 + e t ² ( s L ± e 3 t sin 6 t ² ( s ) ± t 3 ² ( s L ± e t ² ( s ) = 6 ( s 3) 2 +6 2 3! s 3+1 + 1 s 1 = 6 ( s 3) 2 +36 6 s 4 + 1 s 1 , valid for 3. 19. For 5, we have L n t 4 e 5 t e t cos 7 t o ( s L ± t 4 e 5 t ² ( s ) n e t cos 7 t o ( s ) = 4! ( s 5) 4+1 s 1 ( s 1) 2 +( 7) 2 = 24 ( s 5) 5 s 1 ( s 1) 2 +7 . 21. Since the function g 1 ( t ) 1 is continuous on (
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