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2302-practice-final-soln

Therefore the final answer is y t = c t y 1 d t y 2 =

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Unformatted text preview: Therefore the final answer is y ( t ) = C ( t ) y 1 + D ( t ) y- 2 = (- ln t + E ) e − 2 t + parenleftbigg- 1 t + F parenrightbigg te − 2 t . MAP 2302, Fall 2011 — Final Exam Review Problems 5 5 Suppose that the Laplace transform of y ( t ) is given by 2 s 3 + s 2 + 8 s + 6 ( s 2 + 1)( s 2 + 4) . What is y ? Solution: First we need to use partial fractions to split this into two fractions, with denominators s 2 + 1 and s 2 + 4, respectively. Setting 2 s 3 + s 2 + 8 s + 6 ( s 2 + 1)( s 2 + 4) = As + B s 2 + 1 + Cs + D s 2 + 4 , we see by canceling denominators that we need 2 s 3 + s 2 + 8 s + 6 = ( As + B )( s 2 + 4) + ( Cs + D )( s 2 + 1) , which simplifies to 2 s 3 + s 2 + 8 s + 6 = ( A + C ) s 3 + ( B + D ) s 2 + (4 A + C ) s + (4 B + D ) . Therefore we have the system A + C = 2 , B + D = 1 , 4 A + C = 8 , 4 B + D = 6 . From the first and third lines we see that C = 2- A and thus A = 2 and C = 0. From the second and fourth lines we find that B = 5 / 3 and D =- 2 / 3 . Now we have 2 s 3 + s 2 + 8 s + 6 ( s 2 + 1)( s 2 + 4) = 2 s + 5 / 3 s 2 + 1- 2 / 3 s 2 + 4 . In order to compute the inverse transform, we need to do a bit more simplifying: 2 s 3 + s 2 + 8 s + 6 ( s 2 + 1)( s 2 + 4) = 2 parenleftbigg s s 2 + 1 parenrightbigg + 5 3 parenleftbigg 1 s 2 + 1 parenrightbigg- 1 3 parenleftbigg 2 s 2 + 4 parenrightbigg . Now we are nearly done: L − 1 braceleftbigg 2 s 3 + s 2 + 8 s + 6 ( s 2 + 1)( s 2 + 4) bracerightbigg = 2 L − 1 braceleftbigg s s 2 + 1 bracerightbigg + 5 3 L − 1 braceleftbigg 1 s 2 + 1 bracerightbigg- 1 3 L − 1 braceleftbigg 2 s 2 + 4 bracerightbigg , = 2cos t + 5 3 sin t- 1 3 sin 2 t. MAP 2302, Fall 2011 — Final Exam Review Problems 6 6 Suppose that f ( t ) is periodic with period 2 and that f ( t ) = e − t for 0 < t < 1 and f ( t ) = 1 for 1 < t < 2. Compute the Laplace transform of f ( t ). Solution: First define f T ( t ) to be f throughout its first period: f T ( t ) = braceleftbigg e − t < t < 1 , 1 1 < t < 2 . The period of f is T = 2, so we know from the formula sheet that L { f } = L { f T } 1- e − 2 s . To compute L { f T } , we need to express f T in terms of Heaviside functions: f T ( t ) = e − t- u ( t- 1) e − t bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright turn off + u ( t- 1) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright turn on + u ( t- 2) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright turn off . Therefore we see that L { f T } = L braceleftbig e − t bracerightbig- e − s L braceleftBig e − ( t +1) bracerightBig + e − s L { 1 } - e − 2 s L { 1 } , = L braceleftbig e − t bracerightbig- e − s 1 e L braceleftbig e − t bracerightbig + e − s L { 1 } - e − 2 s L { 1 } , = 1 s + 1- e − s 1 e 1 s + 1 + e − s 1 s- e − 2 s 1 s , = 1- e − s − 1 s + 1 + e − s- e − 2 s s ....
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Therefore the final answer is y t = C t y 1 D t y 2 = ln t...

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