Solution first we need to use partial fractions to

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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 , = 2 cos t + 5 3 sin t - 1 3 sin 2 t.
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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 0 < 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 . Our final answer is therefore L { f } = 1 e - s - 1 s +1 + e - s e - 2 s s 1 - e 2 s . 7 Find the first six terms in the power series expansion at x = 0 for the general solution to the differential equation ( x 2 + 1) y ′′ + y = 0. Solution: We need to find the first six terms, so we make the following substitutions y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + . . . , y = a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + 5 a 5 x 4 + . . . , y ′′ = 2 a 2 + 6 a 3 x + 12 a 4 x 2 + 20 a 5 x 3 + . . . ,
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MAP 2302, Fall 2011 — Final Exam Review Problems 7 Now we make the substitution: ( x 2 +1) y ′′ + y = ( x 2 +1)(2 a 2 +6 a 3 x +12 a 4 x 2 +20 a 5 x 3 + . . . )+( a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + . . . ) .
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