L y n s s n y s s n 1 y 0 s n 2 y 0 s y n 2 0 y n 1 0

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. . L [ y ( n ) ]( s ) = s n Y ( s ) s n 1 y (0) s n 2 y (0) − · · · − s y ( n 2) (0) y ( n 1) (0) . (8.9) Application to IVPs. Suppose that y ( t ) is the solution of the initial-value problem 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 . It can be shown that if f ( t ) is piecewise continuous over [0 , ) and is of exponential order as t → ∞ then y ( t ) is n -times differentiable and that it and its first n 1 derivatives are of exponential order as t → ∞ . You can thereby use (8.9) to find Y ( s ) = L [ y ]( s ) in terms of the inital data y 0 , y 1 , · · · , y n 1 , and the Laplace transform of the forcing, F ( s ) = L [ f ]( s ). Indeed, the fact L is a linear operator implies that the Laplace transform of the initial-value problem is L [D n y ] + a 1 L [D n 1 y ] + · · · + a n 1 L [D y ] + a n L [ y ] = L [ f ] . If we use (8.9) then we find p ( s ) Y ( s ) = q ( s ) + F ( s ) , where p ( s ) is the characteristic polynomial p ( s ) = s n + a 1 s n 1 + · · · + a n 1 s + a n . and q ( s ) is the polynomial given in terms of the initial data by q ( s ) = ( s n 1 + a 1 s n 2 + · · · + a n 2 s + a n 1 ) y 0 + ( s n 2 + a 1 s n 3 + · · · + a n 3 s + a n 2 ) y 1 . . . + ( s 2 + a 1 s + a 2 ) y n 3 + ( s + a 1 ) y n 2 + y n 1 . You therefore find that Y ( s ) = q ( s ) + F ( s ) p ( s ) . (8.10)
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11 Later we will see how to determine y ( t ) from Y ( s ), but first we will illustrate how to compute Y ( s ). The hardest part of doing this is often computing F ( s ) = L [ f ]( s ) in (8.10). Usually f ( t ) can be expressed as a combination of the basic forms whose Laplace transform we have already computed. Some of these basic forms are L [ t n ]( s ) = n ! s n +1 L [cos( bt )]( s ) = s s 2 + b 2 L [sin( bt )]( s ) = b s 2 + b 2 L [ e at f ( t )]( s ) = F ( s a ) L [ t n f ( t )]( s ) = ( 1) n F ( n ) ( s ) L [ u ( t c ) f ( t c )]( s ) = e cs F ( s ) L [ e at t n ]( s ) = n ! ( s a ) n +1 L [ e at cos( bt )]( s ) = s a ( s a ) 2 + b 2 L [ e at sin( bt )]( s ) = b ( s a ) 2 + b 2 for s > 0 , for s > 0 , for s > 0 , where F ( s ) = L [ f ( t )]( s ) , where F ( s ) = L [ f ( t )]( s ) , where F ( s ) = L [ f ( t )]( s ) and u ( t ) is the unit step function , for s > a , for s > a , for s > a . (8.11) One can use these to build up a much longer table of basic forms such as the one given in the book. However, the above table contains all the forms you really need. In fact, you can argue the first three entries are redundent because they follow by setting a = 0 in the last three. Alternatively, you can argue that the last three entries are redundent because they follow immediately from the first three and the fourth. On exams you will be given a table that includes at least the last six entries above, so there is no need to memorize this table. However, you should learn how to use it efficiently. Example. Find the Laplace transform Y ( s ) of the solution y ( t ) of the initial-value problem y 2 y = e 5 t , y (0) = 3 . By setting a = 5 and n = 0 in the seventh entry of table (8.11) you see that L [ e 5 t ]( s ) = 1 / ( s 5). The Laplace transform of the initial-value problem therefore is L [ y ]( s ) 2 L [ y ]( s ) = L [ e 5 t ]( s ) = 1 s 5 ,
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12 where we see from (8.9) that L [ y ]( s ) = Y ( s ) , L [ y ]( s ) = s Y ( s ) y (0) = s Y ( s ) 3 .
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