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lecture_23 (dragged) 2

# lecture_23 (dragged) 2 - MA 36600 LECTURE NOTES WEDNESDAY...

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MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11 3 we have L [ y ] = L m i =1 c i y i = m i =1 c i L [ y i ] = m i =1 c i · 0 = 0 . Hence y = y ( t ) is also a solution to the homogeneous problem. As a consequence, we show that the general solution to L [ y ] = G ( t ) is y ( t ) = u ( t ) + Y ( t ), where L [ u ] = 0 and L [ Y ] = G ( t ). To see why, let Y = Y ( t ) denote a particular solution, and y = y ( t ) be any solution to L [ y ] = G ( t ). If we define u ( t ) = y ( t ) Y ( t ), then L [ u ] = L [ y ] L [ Y ] = G ( t ) G ( t ) = 0 . It su ces then to find the general solution to homogeneous equation L [ y ] = 0. Linear Independence. We discuss when we have the general solution to L [ y ] = 0. Consider the initial value problem n i =1 P n j ( t ) y ( j ) = G ( t ) where y ( j 1) ( t 0 ) = y ( j 1) 0 , i = 1 , 2 , . . . , n. Say that { y 1 , y 2 , . . . , y m } is a set of functions satisfying L [ y i ] = 0, and consider the linear combination y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + · · · + c m y m ( t ) We know that L [ y ] = 0. In order for y = y ( t ) to be the general solution, we need to find constants c i such that y ( j 1) ( t 0 ) = y ( j 1) 0 . We may di ff erentiate this expression then evaluate at t = t 0 : y ( t 0 ) = c 1 y 1 (
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