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lecture_24

# lecture_24 - MA 36600 LECTURE NOTES FRIDAY MARCH 13 Higher...

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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 X 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 X 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 6 = 0 . (3) Y = Y ( t ) is a solution to the nonhomogeneous equation n X j =0 P n - j ( t ) Y ( j ) = G ( t ) . Then there exist constants c i such that the solution to the differential equation is in the form y ( t ) = n X 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 it suffices to find constants c i such that y ( t ) - Y ( t ) = u ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + · · · + c n y n ( t ) . is the solution to the initial value problem. We may differentiate this expression then evaluate at t = t 0 : u ( t 0 ) = c 1 y 1 ( t 0 ) + c 2 y 2 ( t 0 ) + · · · + c n y n ( t 0 ) u (1) ( t 0 ) = c 1 y (1) 1 ( t 0 ) + c 2 y (1) 2 ( t 0 ) + · · · + c n y (1) n ( t 0 ) u (2) ( t 0 ) = c 1 y (2) 1 ( t 0 ) + c 2 y (2) 2 ( t 0 ) + · · · + c n y (2) n ( t 0 ) . . . u ( n - 1) ( t 0 ) = c 1 y ( n - 1) 1 ( t 0 ) + c 2 y ( n - 1) 2 ( t 0 ) + · · · + c n y ( n - 1) n ( t 0 ) 1

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2 MA 36600 LECTURE NOTES: FRIDAY, MARCH 13 We write this using the product of matrices:
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lecture_24 - MA 36600 LECTURE NOTES FRIDAY MARCH 13 Higher...

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