hw7key - AMATH 351 Homework 7 Due Aug 12, 2009 Section6.1...

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Unformatted text preview: AMATH 351 Homework 7 Due Aug 12, 2009 Section6.1 13, 19 Section6.2 4, 7, 11, 16, 23 Section 6.1 13. This was the example showed in class. 19. L t 2 sin at Since L{ sin at } = a s 2 + a 2 , s > , according to the last row of the table on page 319, L t 2 sin at = L n (- t ) 2 sin at o = d 2 ds 2 a s 2 + a 2 = d ds- 2 as ( s 2 + a 2 ) 2 ! =- 2 a ( s 2 + a 2 ) 2 + 8 as 2 ( s 2 + a 2 ) 3 = 2 a ( 3 s 2- a 2 ) ( s 2 + a 2 ) 3 , s > Section 6.2 4. 3 s s 2- s- 6 = 3 s ( s- 3) ( s + 2) = A s- 3 + B s + 2 1 Multiply ( s- 3) ( s + 2) on both sides of the equation, we get A ( s + 2) + B ( s- 3) = 3 s then match the coe cients, ( A + B = 3 2 A- 3 B = 0 = ( A = 9 5 B = 6 5 So L- 1 3 s s 2- s- 6 = L- 1 9 5 1 s- 3 + 6 5 1 s + 2 = 9 5 e 3 t + 6 5 e- 2 t . 7. 2 s + 1 s 2- 2 s + 2 = 2 ( s- 1) + 3 ( s- 1) 2 + 1 = 2 s- 1 ( s- 1) 2 + 1 + 3 1 ( s- 1) 2 + 1 So L- 1 2 s + 1 s 2- 2 s + 2 = 2 L- 1 ( s- 1 ( s- 1) 2 + 1 ) + 3 L- 1 ( 1 ( s- 1) 2 + 1 ) = 2 e t cos t + 3 e t sin t 11....
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hw7key - AMATH 351 Homework 7 Due Aug 12, 2009 Section6.1...

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