HigherLin4

# A n 3 s 2 a n 2 s a n 1 y s n 2 a 1 s n 3 a n 3 s a n

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· · · + a n 3 s 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 , because L [D k g ] = s k G ( s ) for every k = 0, 2, · · · , n 1, and because G ( s ) = 1 /p ( s ), the function y H ( t ) can be expressed in terms of g ( t ) as y H ( t ) = y 0 parenleftBig D n 1 + a 1 D n 2 + · · · + a n 3 D 2 + a n 2 D + a n 1 parenrightBig g ( t ) + y 1 parenleftBig D n 2 + a 1 D n 3 + · · · + a n 3 D + a n 2 parenrightBig g ( t ) . . . + y n 3 parenleftBig D 2 + a 1 D + a 2 parenrightBig g ( t ) + y n 2 ( D + a 1 ) g ( t ) + y n 1 g ( t ) . The natural fundamental set of solutions to the associated homogeneous equation L y = 0 is therefore given by Y 1 ( t ) = parenleftBig D n 1 + a 1 D n 2 + · · · + a n 3 D 2 + a n 2 D + a n 1 parenrightBig g ( t ) , Y 2 ( t ) = parenleftBig D n 2 + a 1 D n 3 + · · · + a n 3 D + a n 2 parenrightBig g ( t ) , . . . Y n 2 ( t ) = parenleftBig D 2 + a 1 D + a 2 parenrightBig g ( t ) , Y n 1 ( t ) = ( D + a 1 ) g ( t ) , Y n ( t ) = g ( t ) . (8.18) The solution of the initial value problem L y = f ( t ) , y (0) = y 0 , y (0) = y 1 , · · · y ( n 2) (0) = y n 2 , y ( n 1) (0) = y n 1 , can then be expressed as y ( t ) = y 0 Y 1 ( t ) + y 1 Y 2 ( t ) + · · · + y n 2 Y n 1 ( t ) + y n 1 Y n ( t ) + ( Y n f )( t ) , where Y 1 ( t ), Y 1 ( t ), · · · , Y n ( t ) is the natural fundamental set of solutions to the associated homogeneous equation and is given in terms of the Green function g ( t ) by (8.18). One can check that W [ Y 1 , Y 2 , · · · , Y n ](0) = 1.
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• Spring '10
• LEVERMORE

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