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10-21-review-soln

# 10-21-review-soln - Friday Dierential Equations Fall 2011...

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Differential Equations — Fall 2011 Friday, 10/21/11 In-class review for Section 7.6 — Solutions 1 Express the piecewise defined function f ( t ) = 2 0 < t < 3 , 0 3 < t < 7 , t 5 7 < t < 9 , sin t t > 9 in a single line in terms of Heaviside functions. Solution: It is just 2 - 2 u ( t - 3) + t 5 u ( t - 7) - t 5 u ( t - 9) + (sin t ) u ( t - 9) . 2 Find the Laplace transform of the solution y to the initial value prob- lem 2 y ′′ - 3 y + y = braceleftbigg t 2 0 < t < 5 , sin t t > 5 y (0) = 1; y (0) = 0 . Solution: To save some writing, let Y ( s ) = L { y } . For the lefthand side, we have L { y } = Y, L { y } = sY - y (0) = sY - 1 , L { y ′′ } = s ( L { y } ) - y (0) = s 2 Y - s. So, we get L braceleftbig 2 y ′′ - 3 y + y bracerightbig = 2( s 2 Y - s ) - 3( sY - 1)+ Y = (2 s 2 - 3 s +1) Y +( - 2 s +3) . To transform the righthand side, we first need to express it in terms of Heaviside functions. We have that braceleftbigg t 2 0 < t < 5 , sin t t > 5 = t 2 - t 2 u ( t - 5) + (sin t ) u ( t - 5) ,

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Differential Equations — Fall 2011 Friday, 10/21/11 so using the rule that L { f ( t ) u ( t - a ) } = e as L { f ( t + a ) } , we get that L braceleftbig t 2 - t 2 u ( t - 5) + (sin t ) u ( t - 5)
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10-21-review-soln - Friday Dierential Equations Fall 2011...

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