It follows that s 2 y s 3 1 s 5 y s 1 s 2 s 5 3 s 2

Info icon This preview shows pages 12–14. Sign up to view the full content.

View Full Document Right Arrow Icon
It follows that ( s 2) Y ( s ) 3 = 1 s 5 , = Y ( s ) = 1 ( s 2)( s 5) + 3 s 2 . Example. Find the Laplace transform Y ( s ) of the solution y ( t ) of the initial-value problem y ′′ 2 y 8 y = 0 , y (0) = 3 , y (0) = 7 . Here there is no forcing. The Laplace transform of the initial-value problem is L [ y ′′ ]( s ) 2 L [ y ]( s ) 8 L [ y ]( s ) = 0 , where we see from (8.9) that L [ y ]( s ) = Y ( s ) , L [ y ]( s ) = s Y ( s ) y (0) = s Y ( s ) 3 , L [ y ′′ ]( s ) = s 2 Y ( s ) s y (0) y (0) = s 2 Y ( s ) 3 s 7 . It follows that ( s 2 2 s 8) Y ( s ) 3 s 1 = 0 , = Y ( s ) = 3 s + 1 s 2 2 s 8 . Example. Find the Laplace transform Y ( s ) of the solution y ( t ) of the initial-value problem y ′′ + 4 y = sin(3 t ) , y (0) = y (0) = 0 . By setting b = 3 in the third entry of table (8.11) you see that L [sin(3 t )]( s ) = 3 / ( s 2 + 9). The Laplace transform of the initial-value problem therefore is L [ y ′′ ]( s ) + 4 L [ y ]( s ) = L [sin(3 t )]( s ) = 3 s 2 + 3 2 = 3 s 2 + 9 , where we see from (8.9) that L [ y ]( s ) = Y ( s ) , L [ y ′′ ]( s ) = s 2 Y ( s ) sy (0) y (0) = s 2 Y ( s ) . It follows that ( s 2 + 4) Y ( s ) = 3 s 2 + 9 , = Y ( s ) = 3 ( s 2 + 4)( s 2 + 9) .
Image of page 12

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

View Full Document Right Arrow Icon
13 8.5: Piecewise Defined Forcing. The Laplace transform method can be to solve initial- value problems of the form D n y + a 1 D n 1 y + · · · + a n 1 D y + a n y = f ( t ) , where D = d d t , y (0) = y 0 , y (0) = y 1 , · · · y ( n 1) (0) = y n 1 , where the forcing f ( t ) is piecewise defined over [0 , ) by a list of cases given in the form f ( t ) = f 0 ( t ) for 0 t < c 1 , f 1 ( t ) for c 1 t < c 2 , . . . . . . f m 1 ( t ) for c m 1 t < c m , f m ( t ) for c m t < , where 0 = c 0 < c 1 < c 2 < · · · < c m < . We assume that for each k = 0, 1, · · · , m 1 the function f k is continuous and bounded over [ c k , c k +1 ), while the function f m is continuous over [ c m , ) and is of exponential order as t → ∞ . Here we show how to compute the Laplace transform F ( s ) = L [ f ]( s ) for such a function. There are three steps. The first step is to express f ( t ) in terms of translations of the unit step u ( t ). How this is done becomes clear once you see that for every 0 c < d one has u ( t c ) u ( t d ) = braceleftbigg 1 for c t < d , 0 otherwise . In other words, the function u ( t c ) u ( t d ) is a switch that turns on at t = c and turns off at t = d . This observation allows you to express f ( t ) as f ( t ) = ( u ( t ) u ( t c 1 ) ) f 0 ( t ) + ( u ( t c 1 ) u ( t c 2 ) ) f 1 ( t ) + · · · + ( u ( t c m 1 ) u ( t c m ) ) f m 1 ( t ) + u ( t c m ) f m ( t ) . By grouping terms above that involve the same u ( t c k ), you obtain f ( t ) = f 0 ( t ) + u ( t c 1 ) ( f 1 ( t ) f 0 ( t ) ) + · · · + u ( t c m ) ( f m ( t ) f m 1 ( t ) ) . (8.12) This is the form you want. You can get to it by either repeating the steps given above or by memorizing formula (8.12). If you understand that each term u ( t c k ) ( f k ( t ) f k 1 ( t ) ) appearing in (8.12) simply changes the forcing from f k 1 ( t ) to f k ( t ) at time t = c k then it is not hard to recall.
Image of page 13
Image of page 14
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