# sol8 - s 2 1 s-5 Cs D s-5 s-2 The special value s = 2...

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MAP 2302 Fall-2011 Section 0100 1. Is the function e t 2 t +1 of exponential order? Justify your answer. SOLUTION: Yes. Since t 2 t +1 t 2 t = t , we have e t 2 t +1 e t . 2. Solve the initial value problem by method of Laplace transform. y 00 - 7 y 0 + 10 y = 9 cos t + 7 sin t ; y (0) = 5, y 0 (0) = - 4. SOLUTION: We apply the Laplace transform to the both sides of the equation to obtain ( s 2 Y - 5 s + 4) - 7( sY - 5) + 10 Y = 9 s s 2 + 1 + 7 s 2 + 1 . Solve it for Y : Y = (5 s - 39)( s 2 + 1) + 9 s + 7 ( s 2 - 7 s + 10)( s 2 + 1) . Find the roots: s 2 - 7 s +10 = ( s - 2)( s - 5). Use the partial fraction method (5 s - 39)( s 2 + 1) + 9 s + 7 ( s - 2)( s - 5)( s 2 + 1) = A s - 5 + B s - 2 + Cs + D s 2 + 1 . Then (5 s - 39)( s 2 +1)+9
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Unformatted text preview: ( s 2 +1)( s-5)+( Cs + D )( s-5)( s-2) . The special value s = 2 implies-15 B =-29 × 5 + 18 + 7 =-120. Hence, B = 8 The value s = 5 implies (26 × 3) A =-14 × 26 + 52 =-12 × 26. Hence A =-4. The value s = 0 implies 8-40 + 10 D =-39 + 7. Hence D = 0. The value s = 1 implies 8-64+4 C = (5-39) × 2+16 =-52. Therefore C = 1. Thus, y = L-1 ( Y ) = L-1 {-4 s-5 } + L-1 { 8 s-2 } + L-1 { s s 2 + 1 } =-4 e 5 t +8 e 2 t +cos t. 1...
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