lecture_24 - MA 36600 LECTURE NOTES: FRIDAY, MARCH 13...

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Unformatted text preview: 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 ) = y ( j- 1) , 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 : u ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + + c n y n ( t ) u (1) ( t ) = c 1 y (1) 1 ( t ) + c 2 y (1) 2 ( t ) + + c n y (1) n ( t ) u (2) ( t ) =...
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue University-West Lafayette.

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lecture_24 - MA 36600 LECTURE NOTES: FRIDAY, MARCH 13...

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