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# lecture_24 (dragged) - MA 36600 LECTURE NOTES FRIDAY MARCH...

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MA 36600 LECTURE NOTES: FRIDAY, MARCH 13 Higher Order Linear Equations Fundamental Set of Solutions. Consider the initial value problem n j =0 P n j ( t ) y ( j ) = G ( t ) where y ( j 1) ( t 0 ) = y ( j 1) 0 , i = 1 , 2 , . . . , n. Say that we can find ( n + 1) functions y 1 = y 1 ( t ) , . . . , y n ( t ) and Y = Y ( t ) such that (1) Each y i = y i ( t ) is a solution to the homogeneous equation n j =0 P n j ( t ) y ( j ) i = 0 for i = 1 , 2 , . . . , n . (2) The n × n determinant is nonzero: W y 1 , y 2 , . . . , y n ( t ) = y 1 ( t ) y 2 ( t ) · · · y n ( t ) y (1) 1 ( t ) y (1) 2 ( t ) · · · y (1) n ( t ) . . . . . . . . . . . . y ( n 1) 1 ( t ) y ( n 1) 2 ( t ) · · · y ( n 1) n ( t ) = y ( j 1) i = 0 . (3) Y = Y ( t ) is a solution to the nonhomogeneous equation n j =0 P n j ( t ) Y ( j ) = G ( t ) . Then there exist constants c i such that the solution to the di ff erential equation is in the form y ( t ) = n i =1 c i y i ( t ) + Y ( t ) . For this reason, we call { y 1 , y 2 , . . . , y n } a fundamental set of solutions. We explain why. We know that u ( t ) = y ( t ) Y ( t ) is a solution to the homogeneous equation by (3), so
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