10-14-review-soln - Differential Equations — Fall 2011...

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Unformatted text preview: Differential Equations — Fall 2011 Friday, 10/14/11 In-class review for Sections 7.2–7.5 — Solutions 1 Find the Laplace transform of the solution y to the initial value prob- lem y + 2 y + 2 y = 1 for 0 ≤ t ≤ 7, t for t > 7. ; y (0) = 2 , y (0) = 1 . Solution: Let Y = L { y } . For the lefthand side, we have the following transforms: L { y } = Y, L { y } = sY- y (0) = sY- 2 , L { y } = s L { y } - y (0) = s 2 Y- 2 s- 1 . For the righthand side, we just have to do the integral: L 1 for 0 ≤ t ≤ 7, t for t > 7. = 7 e- st dt + ∞ 7 te- st dt, =- 1 s e- st 7 + lim N →∞- t s e- st- 1 s 2 e- st N 7 , =- e- 7 s s + 1 s + lim N →∞- Ne- Ns s- e- Ns s 2 + 7 e- 7 s s + e- 7 s s 2 , = 6 e- 7 s s + 1 s + e- 7 s s 2 . Now we need to put this all together. Taking the Laplace transform of both sides, we have L y + 2 y + 2 y = L 1 for 0 ≤ t ≤ 7, t for t > 7, so ( s 2 Y- 2 s- 1 ) + 2 ( sY- 2) + 2 Y = 6 e- 7 s s + 1 s + e- 7 s s 2 ....
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This note was uploaded on 12/10/2011 for the course MAP 2302 taught by Professor Tuncer during the Fall '08 term at University of Florida.

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10-14-review-soln - Differential Equations — Fall 2011...

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