HigherLin4

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

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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) .

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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.
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