2302-practice-mid3-soln

2302-practice-mid3-soln - MAP 2302, Fall 2010 Midterm 3...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAP 2302, Fall 2010 Midterm 3 Review Problems 1 The exam will cover sections 7.2, 7.3, 7.4, 7.5, 7.6, and 7.8. All topics from this review sheet or from the suggested exercises are fair game. 1 Solve for L { y } given the following initial value problems. a. y - 4 y + 8 y = e 2 t cos 3 t ; y (0) = 1; y (0) = 3. Solution: Let Y ( s ) = L { y } . We have L { y } = Y, L braceleftbig y bracerightbig = sY- y (0) = sY- 1 , L braceleftbig y bracerightbig = s ( L { y } )- y (0) = s 2 Y- s- 3 , so L braceleftbig y - 4 y + 8 y bracerightbig = ( s 2 Y- s- 3)- 4( sY- 1) + 8 Y = ( s 2- 4 s + 8) Y + (- s + 1) . For the righthand side, we first see that L { cos 3 t } = s s 2 + 9 , so using the rule L braceleftbig e at f ( t ) bracerightbig = F ( s- a ), where F ( s ) = L { f } , we see that L braceleftbig e 2 t cos 3 t bracerightbig = s- 2 ( s- 2) 2 + 9 . Putting these two together and solving for Y , we get Y ( s ) = s 2 ( s 2) 2 +9 + s- 1 s 2- 4 s + 8 . b. y + 2 y - 3 y = e t + t + 1; y (0) = 9; y (0) =- 3. Solution: Letting Y ( s ) = L { y } we have L { y } = Y, L braceleftbig y bracerightbig = sY- y (0) = sY- 9 , L braceleftbig y bracerightbig = s ( L { y } )- y (0) = s 2 Y- 9 s + 3 , MAP 2302, Fall 2010 Midterm 3 Review Problems 2 so L braceleftbig y + 2 y - 3 y bracerightbig = ( s 2 Y- 9 s + 3) + 2( sY- 9)- 3 Y = ( s 2 + 2 s- 3) Y + (- 9 s- 15) . For the righthand side, we have L braceleftbig e t + t + 1 bracerightbig = 1 s- 1 + 1 s 2 + 1 s . Solving for Y , we get Y = 1 s 1 + 1 s 2 + 1 s + 9 s + 15 s 2 + 2 s- 3 . c. y - 4 y = braceleftbigg sin t < t < ,- sin t t > . ; y (0) = y (0) = 0. Solution: Letting Y ( s ) = L { y } we have L { y } = Y, L braceleftbig y bracerightbig = sY- y (0) = sY, L braceleftbig y bracerightbig = s ( L { y } )- y (0) = s 2 Y, so the lefthand side is transformed to L braceleftbig y - 4 y bracerightbig = s 2 Y- 4 Y = ( s 2- 4) Y. In order to transform the righthand side, we first convert it into Heaviside functions: braceleftbigg sin t < t < ,- sin t t > ....
View Full Document

This note was uploaded on 12/10/2011 for the course MAP 2302 taught by Professor Tuncer during the Fall '08 term at University of Florida.

Page1 / 10

2302-practice-mid3-soln - MAP 2302, Fall 2010 Midterm 3...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online