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# lecture_23 - MA 36600 LECTURE NOTES WEDNESDAY MARCH 11...

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MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11 Higher Order Linear Equations Review. Recall that a second order linear equation is in the form a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = f ( t ) . We found that the general solution is in the form y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Y ( t ) where { y 1 ( t ) , y 2 ( t ) } is a fundamental set of solutions to the homogeneous equation a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = 0 and Y = Y ( t ) is a particular solution to the nonhomogeneous equation a ( t ) Y 00 + b ( t ) Y 0 + c ( t ) Y = f ( t ) . We give a different way to view this. Define the following operator: L [ y ] = a ( t ) d 2 y dt 2 + b ( t ) dy dt + c ( t ) y. This is a linear operator , i.e., given functions f = f ( t ) and g = g ( t ) as well as constants c 1 and c 2 , we have L c 1 f + c 2 y = a ( t ) d 2 dt 2 c 1 f ( t ) + c 2 g ( t ) + b ( t ) d dt c 1 f ( t ) + c 2 g ( t ) + c ( t ) c 1 f ( t ) + c 2 g ( t ) = a ( t ) c 1 d 2 f dt 2 + c 2 d 2 g dt 2 + b ( t ) c 1 df dt + c 2 dg dt + c ( t ) c 1 f + c 2 g = c 1 a ( t ) d 2 f dt 2 + b ( t ) df dt + c ( t ) f + c 2 a ( t ) d 2 g dt 2 + b ( t ) dg dt + c ( t ) g = c 1 L [ f ] + c 2 L [ g ] . We wish to find all functions y = y ( t ) such that L [ y ] = f ( t ). In a sense, we would like to compute y ( t ) = L - 1 [ f ]. We do this in two steps: First we find functions y 1 = y 1 ( t ) and y 2 = y 2 ( t ) such that L [ y 1 ] = L [ y 2 ] = 0 and W ( y 1 , y 2 ) ( t ) = y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) = y 1 ( t ) y 0 2 ( t ) - y 0 1 ( t ) y 2 ( t ) 6 = 0 . Then L [ y ] = 0 if and only if y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) . Similarly, L [ y ] = f ( t ) if and only if y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Z t f ( τ ) a ( τ ) y 1 ( τ ) y 2 ( t ) - y 1 ( t ) y 2 ( τ ) y 1 ( τ ) y 0 2 ( τ ) - y 0 1 ( τ ) y 2 ( τ ) dτ.

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lecture_23 - MA 36600 LECTURE NOTES WEDNESDAY MARCH 11...

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