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Unformatted text preview: Differential Equations — Fall 2011 Friday, 10/14/11 Inclass 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.
 Fall '08
 TUNCER

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