2302-practice-final-soln

Solution first we need to use partial fractions to

Info icon This preview shows pages 5–8. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 + . . . ,
Image of page 6
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 + . . . ) .
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern